I was interviewing with a options trading...

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I was interviewing with a options trading firm and was asked this question:

>There is an asset whose price goes up 30% with 50% probability

>It goes down 25% with 50% probability

>After thousands of iterations of these prices changes, what is the expected return?

I don't understand...the expected return in a single time frame is positive, but after running simulations on my own, I know that this eventually falls to 0 with many iterations.

Why is this?

>>

50/50, it either goes up or it doesn't

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>>7635512

0.75*1.3 = 0.975

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>>7635512

Why would this not just be expected value?

(0.5 * 1.3x) + (0.5 * 0.75x) = 1.025x where x is the price of the asset

I'm not very good with stats so I'm not being a smartass. I'm legitimately asking why it wouldn't just be that.

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>>7635534

the expected value is 0.5(1.3) + 0.5(0.75) = 1.025, no?

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>>7635558

Yeah exactly, that's my question. After running this simulation many times, x always goes to 0, instead of the 1.025x that I expect.

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>>7635512

30 and 25 are too close together being that the chance of it moving in either direction is the same.

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>>7635565

For ease of communication, let the price of that asset be equal to x. Then, there are two cases, either it goes up by 30% first or it goes down by 25% first. For the first case, the new price will be 1.3x, and if it goes down next, it will now be .975x. Going to the second case now, .75x and .975x again. As it should be plainly evident now, if both changes have equal probability, the overall price will tend downwards and eventually reach 0 with infinite time.

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>>7635563

Yes. But that's from a single iteration. So you might then think that you can just take 1.025^inf and be done with it, but you'd be wrong.

The right way to look at it is to say that you start with some capital C, and then you multiply that with 0.75, 1.3, 0.75, 0.75, 1.3, and so on. So in the end you get C * 0.75^n * 1.3^m, and letting n, m go to inf you get C*(0.75*1.3)^inf = 0.

The reason why repeatedly applying the expected value/return doesn't work, is that when the capital C goes down the absolute value of the expeted return also goes down, even thought the relative expected return stays the same.

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>>7635590

I'm not so sure. For large N, you'd expect N/2 "wins" and N/2 "losses", but would the answer simply be (1.3)^(N/2) (0.75)^(N/2) = (0.975)^(N/2) ? I'm wondering if there is non-linearity that makes the problem more difficult, because it doesn't allow for the passing of E() across terms so easily.

Not saying it's wrong, just have doubts.

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>>7635907

Tried working it out...

Prob( n wins in N trials) = C(n,k) p^n (1-p)^(N-k)

Rate of return for n wins in N trials = ((1+r)^n (1+s)^(N-k) - 1)

E = sum_k=0^N C(n,k) ((1+r)^n (1+s)^(N-k) - 1) p^n (1-p)^(N-k)

= -1 + sum_k=0^N C(n,k) ((1+r)^n (1+s)^(N-k) ) p^n (1-p)^(N-k)

= -1 + sum_k=0^N C(n,k) (p(1+r))^n ((1-p)*(1+s))^(N-k)

= -1 + ( (p(1+r)) + (1-p)*(1+s) )^N

p=1/2, r=0.30, s= -0.25

E = 1.025^N -1

As N-> infty, expected rate of return goes to infinity.

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>>7635980

sorry, have p^n in there rather than p^k, but I think the result is right.

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>>7635983

Lrn2 [math] \LaTeX [/math]

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>>7635907

>Not saying it's wrong, just have doubts.

I have doubts too, and yet my little Perl script says I'm wrong.

>>7635588

OK, so going up by 30% , then down by 25%, I'm going down more than up because 1.3 is more than 1 * 1.25.

Wow, that's WAY counter-intuitive.

How can I take this idea to Vegas and retire at the ripe young age of 50?

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>>7636009

>How can I take this idea to Vegas and retire at the ripe young age of 50?

Use it to scam people, obviously. The best way to get rich in Vegas is to own a casino.

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>>7636009

>counter-intuitive.

I think it's wrong. He's calculating the expected final balance at the end of N rounds. The question was the expected rate of return. Not the rate of the expected return.

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Losing 25% brings a number down more than gaining 30%

Its that simple.

Look:

100 × 1.3 = 130

130 × .75 = 97.5

97.5 is less than where you started, at 100.

Also, Hitler did nothing wrong.

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>>7635980

[math]E = \sum_{k=0}^N \binom{n}{k} ((1+r)^k (1+s)^{N-k} - 1) p^k (1-p)^{N-k}[/math]

[math] = -1 + \sum_{k=0}^N \binom{n}{k} \Big((1+r)\,p\Big)^k \Big((1+s)\,(1-p)\Big)^{N-k}[/math]

[math] = -1 + \Big((1+r)\, p + (1+s)\,(1-p)\Big)^N = 1.025^N -1 [/math]

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You guys are seriously dumb.

sum_(k=0)^n (1.3^k 0.75^(n-k))/2^n = (26^(n+1)-15^(n+1))/(11*40^n)

[math]\sum_{k=0}^N 1.3^k 0.75^{N-k} / 2^N

= (26^{n+1} - 15^{n+1}) / (11*40^n)[/math]

ROI is simply

[math](26^{n+1} - 15^{n+1}) / (11*40^n) - 1[/math]

This approaches -1 as n-> infinity

In other words, you lose your money.

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Use logarithm of return instead of percent return, this gives you a sum instead of a product and allows you to compute the expected return with

0.5*ln(0.75)+0.5*ln(1.3) = -0.012 and hence exp(-0.012*N)=0 as N tends to infinity.

Useful reading if you're serious about the job:

https://quantivity.wordpress.com/2011/02/21/why-log-returns/

>>

>>

Let A*(1.3^n)(0.75^k) be the value of a random permutation, where A is the initial Asset.

The median Asset after (n+k) changes is A*1.3^((n+k)/2)*(0.75^(n+k)/2)=A*0.975^((n+k)/2)<A

however the mathmatically the expected value would be Ai(n+k+1)=Ai(n+k)(1.3+0.75)/2>Ai(n+k)

so basically just split you asset in infinetly small subassets an you make 2.5% gains per change

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>>7635588

there are four cases, not two

>>

please provide interval.

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>>7636052

but if this is correct, you will make infinite dollars?

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>>7635512

I made a short program in octave and I get huge numbers for 500-1k iterations, I start getting numbers below 1 at 2k iterations, most of the numbers are below 1 at 2500 and at 3k+ they're all minute... why would you get big numbers at the start but then it falls off? I'm taking averages over a hundred sets of iterations. Starting capital is 1. I'm pretty rusty with programming but this seems right?

k=0;

n=[];

while (k<200)

i=rand(2500,1);

c=1;

for m=1:length(i)

if (i(m) < 0.5)

c=0.75*c;

elseif (i(m) > 0.5)

c=1.3*c;

endif

endfor

k++;

n=[n,c];

endwhile

a=mean(n)

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>>7635563

No, you have to rehash in logarithms so you can use addition

log(1.3)=0.113943352

log(.75)=-0.124938737

E[log(value)]= (0.113943352-0.124938737)/2=-0.0054976925

E[value]=10^-0.0054976925=0.987420882

E[value]^1000= 3.17912387*10^-6

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How can it reach 0 if it can only go down 25%?

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>>7640649

Yes, because E(exp(X)) =exp(E(X)).

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>>7635512

Why do you guys make it so complicated?

if it was 33.3..% then the expectation price would have been 1 since (4/3)*(3/4) = 1,

since 30% < 33.33% the price will go down to 0.

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>>7640693

rounding limitiations on their computers.

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>>7641388

The problem is a little vague

(1+1/3) 1/2 + (1-1/4) (1/2) = 25/24

If we do 1000 experiments, and calculate the rate of return for each, and average those rates of return, I think you get

>>7636052

But if you do 1000 experiments, figure out the final balance for each, average those, then figure the rate of return from that, you may get something different.

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>>7641388

This guy's got the right intuition. Especially for an interview situation

>>

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>>7641637

that's because your first 25/24 result is misguided, you don't know what you're calculating. the problem is very clear.

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Dont take it to infinity, take it to 8000

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Would it matter if the loss or gain happened first?

If you lost 75% of your initial investment on the first try, would that still bring the return down significantly?

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This is a typical martingale problem.

Here, the price of the asset at time n is a positive submartingale. Its expectation goes to infinity (this is easy) but X_n goes to 0 almost everywhere. I'm not quite sure how to prove this but this might be done with easy probablistic techniques like Borel-Cantelli

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>>7635514

> that guy

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>>7642003

X_n is never zero for finite n, what are you smoking

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>>7642021

X_n is never zero but goes to zero almost everywhere. I never said X_n was zero, I said its limit was zero.

Just go back to your high-school calculus class please.

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The answer they provide is only correct if

1.

you assume that when the stock is worth 0 it is impossible for it to gain value again. (is this reasonable? I mean let's say the stock is for a particular dog's shit. No one wants to buy dog shit now but it's possible in the future people discover that this dog shit has special properties so then the stock becomes worth something again)

2. that you interpret

>There is an asset whose price goes up 30% with 50% probability

>It goes down 25% with 50% probability

to mean

>There is an asset whose price goes up 30% of the original price with 50% probability

>It goes down 25% of the original price with 50% probability

linguistically the question is actually saying that it's current price goes up 30% or down 25%.

But when you do assume both these things and answer a significantly different question to the one asked, then it is easy to see that eventually there will be a run of enough losses in a row to reach bankruptcy.

in markov chains you call this recurrence. The value is guaranteed to go back to 0 eventually. (guaranteed to go to any state in the state space if it were allowed to continue after hitting 0.

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>>7642025

>almost everywhere

This has no meaning for the sequence X_n.

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>>7641964

>expected value

You are saying "Expected return" has nothing to do with expected value?

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>>7642055

just wait 2 minutes, I'm writing a detailed solution.

Also, don't bother answering if you don't know basic probability concepts :https://en.wikipedia.org/wiki/Almost_everywhere

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>>7635590

>Yes. But that's from a single iteration. So you might then think that you can just take 1.025^inf and be done with it, but you'd be wrong.

no, he isn't wrong you idiot.

do you not know what a fucking expected value is?

The question is asking for nothing more than the expected value of the stock after thousands of goes.

It is NOT asking "what is the probability that the stock is worth close to 0, which may or may not be high".

Just by looking at 1 stage, you can tell that for every n, the outcomes where the stock has mostly gone up in value will more than make up for the outcomes where the stock has gone down in value.

the expected return is the expected value minus the initial so 1.025^n - 1 which eventually gets very big.

Christ so many retards in this thread

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>>7642067

This is a discrete problem, why use measure spaces?

>>

OP, here is the detailed solution. Next time you'll get the job.

American people are so fucking retarded at maths...

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>>7636219

this is absolute meaningless garbage. if you're trying to do the sum of all the possible price changes you forgot the binomial coefficients you retard.

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>>7642080

hey, it's not measure spaces, it's basic probability.

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>>7642082

neat.

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>>7642075

>the outcomes where the stock has mostly gone up in value will more than make up for the outcomes where the stock has gone down in value.

1. Start with 100 dollars

2a (up 30%): $130

3a (down 25%): $97.5

1. Start with 100 dollars

2b (down 25%): $75

3b (up 30%): $97.5

Going up in value and going down in value are equally likely. But if a stock goes down in value 25%, it would need to go up 33% to get back to even (or if a stock goes up 30%, it would take a ~23% loss to get back to even). Each downtick loses more money than is gained by an uptick at that time.

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>>7642086

>probability has nothing to do with measure spaces

wew lad

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>>7642091

Wow great. I guess by the same reasoning if I lose 1 dollar 99% of the time but win 100 dollars 1% of the time then I shouldn't play the game.

learn what an expected value is. Christ.

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>>7642095

I never said probability had nothing to do with measure spaces. I said there is no need to have a deep understanding on measure spaces to solve basic problems like this one.

tl;dr knowing Radon-Nikodym is not required to solve do high-school proba.

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>>7642095

>I'm trying to appear smart when really I'm just dropping buzzwords about things I do not understand

saying this problem is about or uses or requires measure theory is like saying that solving the equation x-4=5 uses ring theory

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>>7642103

>solving the equation x-4=5 uses ring theory

basically you use the ring properties of Z

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>>7642103

'x-4=5' would be group theory, not ring theory

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>>7642082

>ln(random variable) = E(ln(random variable))

Nope.

by that logic Xn also equals X0 multipled by 1.05^n

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>>7642113

>not understanding basic stats

Are you a burger?

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>>7642113

are you retarded ? Do you know the law of large numbers ?

>>

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>>7642113

I hope you're no older than 15

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>>7642082

>> 1/2 (1.3 + 0.75) = 1.05

Into the trash it goes...

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>>7642118

yeah I was just kidding, obviously you don't need ring theory to solve this nor you don't need measure theory to solve OP's problem.

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>>7642097

How the fuck do you think that game is similar to the one OP laid out?

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>>7642121

my mistake, it was 1.025, but this changes absolutely nothing to the problem. The point is that this quantity is >1.

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>>7635603

this was actually a very interesting problem, I never thought about having to choose between geometric and arithmetic averaging when computing expectencies.

Thanks for bringing this up OP.

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>>7642082

beautiful LaTeX btw.

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>>7642114

I gather that English is not your first language so I'm sorry if you cannot understand my explanation of your error, but you have been remarkably inconsistent.

If you can switch between evaluating a product of random variables and evaluating the exponential of the sum of the expected values of the logs of those random variables at will, then you might as well simply evaluate the product of random variables as the product of the expected values of those random variables.

You're implicitly replacing ln(random variable) with E(ln(random variable)) in order to arrive at x0exp(-infinity). If you're going to do that then you might as well just say that Xn=x0 random variable 1...random variable n -> x0 1.05^infinity

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>>7642123

Because I actually read the question

>what is the expected return?

fuck off , retard

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>>7642131

You just don't understand the LLG. Nobody evaluates the exp of the sum of expected values, but only the exp of the sum of the random variables. This sum goes to -infty because of the LLG.

No expectations are taken here.

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>>7642082

>gros nul

rofl

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>>7642131

enrol in a college probability course then go back to this thread in one year

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>>7642133

by law of large numbers, (randomvariable1.randomvariable2. ... .randomvariableN)^(1/N) -> E(randomvariable)

therefore

Xn -> X0 .1.05^infinity = infinity almost everywhere

Furthermore, one can say that the Arithmetic mean is always at least as high as the geometric mean and so by using a geometric mean I have actually hindered myself and still succeeded in what I wanted to show.

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>>7642131

>I'm sorry if you cannot understand my explanation of your error

you're the error, math guy did no mistake.

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>>7642159

No. Your law of large numbers is wrong.

1) the classical statement is additive. The LLG states that

(1/n * (X1 + ... + X_n) goes to E[random variable]

2) you can switch to multiplicative LLG. Then it states that

(X1*...*Xn)^1/n goes to exp E[log of random variable]

3) to go from 1) to 2), you have to take the exp of the log, then apply the classical LLG to the logs. And it is not true that exp E [ ln X] = E[X].

This is why math guy's proof is correct and yours is wrong.

>>

>.the expected return in a single time frame is positive, but after running simulations on my own, I know that this eventually falls to 0 with many iterations.

So a small number of times the asset will achieve an extremely lare value, or else the expected value would not be positive.

If the expected value of a general step is positive then the expected value of the overall product sequence must be positive too

All they were asking you for was the expected return.

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>>7642082

Why do you have X0 as a random variable? The rate of return is obviously independent of the initial investment. And if you do want X0 as a random variable, why is convergence "almost everywhere" and not simply "everywhere"?

And why in the second part are you calculating exp( E(log(X) ) ? What does that have to do with the problem, which is asking for E(X)?

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>>7642204

In asset pricing, the initial value is often a random variable, but if you like you can consider X0 to be a fixed value.

Whether X0 is random or not, the rate of return is obviously independant of X0, this is what I used in the first paragraph when computing the expected value of Xn.

Convergence is almost everywhere as a consequence of the law of large numbers, the conclusion of which is not everywhere but only almost everywhere but if you're not familiar with probability theory, you can consider that "almost everywhere" and "everywhere" have the same meaning, this will not be a problem.

In the second part, I calculate exp E[ln X] because it's the value to wich exp ln(Xn)/n converges.

The problem does NOT ask for E[X], but asks the limit of Xn. The interesting thing is that Xn goes to infinity IN EXPACTATION, but goes to zero (ALMOST) EVERYWHERE.

The difference between convergence in expectation and (almost) everuwhere is subtle and this problem is an illustration of the difference between the 2.

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>>7642204

...

Looking at the second part again, I wonder if you mean "almost surely" ? Then it makes sense to me. X_n goes to zero almost surely, even though the expected rate of return after N periods is 1.025^N-1.

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>>7642288

>The problem does NOT ask for E[X], but asks the limit of Xn

Well, I disagree with this. OP's statement says "what is the EXPECTED return", i.e. the expected value of the return on investment.

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>>7635512

>After thousands of iterations of these prices changes, what is the expected return?

Trick question. The expected return on an option is always 0 because come option expiration the price is 0.

You got tricked OP.

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>>7642298

you got it right.

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>>7635512

>after running simulations on my own, I know that this eventually falls to 0 with many iterations.

Samefag as >>7642305

On a strict asset without time expiration, you'll still see it go down to 0 due to time decay.

>asset is 100

>asset moves up 30% to 130

>asset moves down 25% to 97.5

If you repeat this enough times you will get to 0. Time decay sucks!

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>>7635512

With my admittedly very basic knowledge of finance, I will say this. Draw downs hurt more than gains help. This anon gets it>>7642091. This is why volatility is bad for a portfolio. The more you can decrease volatility, the better your long term compounded returns will be, even if the average return is marginally greater in a more volatile portfolio. I think this was the point of the question you were asked.

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>>7642301

ah ok.

I answered both questions so it's ok.

Expected value goes to infinity, but Xn goes to zero almost surely.

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>>7642330

That doesn't answer the question though. The question asks for the expected return. This is a strictly defined statistical concept and is not 0.

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>>7642334

>The more you can decrease volatility, the better your long term compounded returns will be, even if the average return is marginally greater in a more volatile portfolio.

Not true. 100% stocks beats 60/40.

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>>7642341

The expected return on each iteration or at the end?

rest of chart from >>7642354

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>>7642354

Thats only because recently bonds have had such shit returns for the past 15 years. Decreasing volatility is pretty basic portfolio theory. Thats why all the interest in investments 'non correlated' to the stock markets - hedgefunds, private equity etc

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>>7642374

>Thats only because recently bonds have had such shit returns for the past 15 years.

They've had the biggest bull market in their history, not sure where you are located if bonds have been performing poorl.y

> Decreasing volatility is pretty basic portfolio theory.

That's not to improve returns though. It may improve risk adjusted returns but no total return.

>hedgefunds, private equity etc

It's true there is interest in non-correlated assets but hedge funds and private equity are almost 100% correlated to stocks.

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>>7642384

OK, I obviously have no idea what I'm talking about. Carry on...

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>>7642288

>The interesting thing is that Xn goes to infinity IN EXPACTATION, but goes to zero (ALMOST) EVERYWHERE.

What if you plotted the probability distribution for possible returns on your investment after N time periods? Wouldn't it look like a Gaussian (for large N) with mean 1.025^N-1 ? So for any epsilon>0, would you have Pr( r<epsilon) --> 0 as N -> infty, where r is your return? I picture gaussians moving off to infinity, leaving a little tail at 0... yet X_n --> 0 almost surely.

So not only does E() go to infinity, it seems the whole distribution does as well in a rough sense.

Is that right???!

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>>7642398

It's OK. You have the right idea but just slightly out of context.

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>>7642382

thank you :)

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>>7639652

and get murdered by transaction fees

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>>7642082

how the fuck did you have the idea to take the exp then apply great numbers law ?

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>>7643489

not him, but french education I guess

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>>7642097

It's not a matter of winning $100. That is a fixed value. It's a matter of winning a percentage of what you have left, and that makes all the difference.

It's funny how similar this is to the two envelopes problem, and how people make the same mistaken assumptions.

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>>7642082

Intredasting

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>>7635512

>after running simulations on my own

sounds like a rounding error in your code desu senpai. what language are you using? e.g. if using Java make sure to use float or long

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>>7642082

What does fat E mean?

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>>7642415

>Is that right???!

It's not. If you start with $1, and X_N is the value after N iterations, then X_N is approximately a log-normal distribution. This means that log(X_N) is approximately normal with mean -1/2 N log(40/39) and variance 1/4 N log^2(26/15). So log(X_N) is getting broader and generally moving towards -infinity as N goes up. But X_N itself is getting piled up very quickly around 0, even though mean(X_N) is growing towards infinity.

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>>7644043

expectation. usually it's written with \mathbb, but I like to write probabilities and expectations with boldfont letters.

>>7643489

It was rather naturel indeed : as someone stated earlier, you feel that it will be a multiplicative problem, and the interesting quantity will be pq, not p+q. So getting to the exp/log was natural.

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>>7644183

>expectation. usually it's written with \mathbb, but I like to write probabilities and expectations with boldfont letters.

Tell me about expectation. I understood your LaTeX post except for the part with the fat E. Where can I read more about this stuff? It's very interesting

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>>7644187

It's a very classical concept in probability : it exactly the "mean" of something random, but made rigorous, cf https://en.wikipedia.org/wiki/Expected_value.

Basically, when you have a random variable which can take the value a with prob. q and the value b with prob 1-q, then the "mean" is a*q + b*(1-q). The expected value generalises thsis.

The general theory is generally accessible to high-school students, but if you want the rigorous side, it's more something for undergraduates.

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>>7644191

I am a Math undergrad but I want to learn more and more and more than the curriculum currently provides. Any books you recommend? Preferably one with exercises in it?

>>

>>7644199

Athreya Lahiri - Measure Theory and Probability Theory

(if you're lying about being a math undergrad, then DONT use this)

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>>7635512

I think a simplified babby way of looking at it is by examining an asset whose price goes down by 50% with 50% probability.

For this transaction to balance out in the long run, you need the asset to go up by 200% with 50% probability. This is because 2 is the inverse of 1/2.

In your case, 13/10 is not the inverse of 3/4. You would need around 4/3 -- that is, 33.33% for it to balance out in the long run.

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>>7644208

Wait, I am not an American. Being an "undergrad" means you're IN undergrad, right? So studying for your bachelor's degree?

Because I don't have an undergrad degree yet.

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>>7644219

>Being an "undergrad" means you're IN undergrad, right?

Yes. When you receive your degree, you are a graduate.

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>>7644221

And what are you called when you get your degree from grad school, or as we like to call it, a "MSc"?

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>>7644199

there are HUNDREDS of books on this. Go to your university library and check "measure theory" or "probability theory", take the one you'll feel comfortable with.

One of my favourite books on this topic is Jacod-Protter, I don't remember the name but it's something like "probability essentials". It has everyhting you'll need in probability theory, introducing what you need in measure theory but not going too far on this way. It has a wealth of exercices (more than 300) and it's quite short (200-300 pages).

The end it quite advanced, with martingale techniques.

Have fun !

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>>7644238

There's undergrad student (in undergrad), graduate student (in graduate studies), and then once you're done whatever title you hold (MSc, PhD, ScD, whatever)

>>

Did this real quick in matlab for those who think it goes to infinity.

Starting value used was 1.

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>>7635512

there was some variation of this problem in the Ross probability book

http://zalsiary.kau.edu.sa/Files/0009120/Files/119387_A_First_Course_in_Probability_8th_Edition.pdf

>>

Bump for interesting thread

>>

75% is 3/4 and 133% is 4/3 so you're only ever going up by less than 4/3.

>>

You retards can't into eigenvalues and eigenvectors

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>>7645889

Holy fuck! How did you picture this?

>>

This is weird, A good /sci/ thread

>>

I am shit at math so I don't know what to do.

>>

>>7642082

complicated answer

>>

>>7645999

and yet the simplest.

How about we ask the real question now: Should you buy such an asset?

>>

>>7646213

I think it would be foolhardy. It shows the danger of relying on a single statistic, like the expected value, in the analysis of possible outcomes. If you look at the distribution as a whole, you can see that it gets smashed down to zero very quickly.

>>

>>7635512

>I don't understand...the expected return

Enjoy your career in the food service industry.

>>

>>7647712

but wouldn't that be an irrational decision?

>>

>>7648294

decreasing marginal utility says "no"

>>

>>7635512

Well consider there's a nonzero chance that your capital will become 0 at one point. It will stay zero from that point on. For example if you have a very unlucky streak of 1 million "down with 25%" it will become close to zero. But your simulations interpret "close to zero" as actual zero, so yeah there's your answer

>>

>>7648907

1.I don't really care about decreasing marginal utility

2.It actually still says yes...

>>

>>7649375

because you don't understand the situation well enough. your decision is the irrational one.

>>

>>7649367

>appeal to faulty approximations

you don't need this at all

>>

>>7649375

If all you care about is expected value, then you have no way to distinguish between investments with the same expected value but vast differences in risk.

>>

>>7649381

>because you don't understand the situation well enough

Oh but I do, that's why I asked the question since it's not so obvious. You on the other hand, might be too limited mentally to understand another point of view than yours, which had its justification disproved.

>>

>>7635512

I would look him straight in the eye and say analysis of this kind is impossible, because the market has complete information. If there were gains to be made from such a prediction, and a prediction were possible, then that would already be reflected in today's price.

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