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I've got a homework question which I think I have solved but am not certain if the answers are correct. The question goes like this: Consider a stock which has a 50% chance of increasing by 80% by the end of the week and a 50% chance of decreasing by 60% after a week. The initial investment amount is $10000.

a) What is the expected value of the investment after 52 weeks?

I used a binomial model for the stock price so that the expected value is $10000*(0.5*1.8 + 0.5*0.4)^{52}$. Is this correct?

b) What is the most likely value your investment will be?

Based on the binomial model again, the value of the investment at the end of 52 weeks can be written as $V(52) = 10000 *1.8^X *0.4^{n-X}$, where $X \sim bin(n,p)$, $n=52$ and $p=0.5$. In this case, is the most likely value found using the mean of $X$; i.e, $np=26$? In other words, my solution is $V(52) = 10000 *1.8^{26} *0.4^{52-26}$.

Would appreciate any advice and/or comments on whether this is the correct way to solve the problem. Thanks!

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It looks right. –  Jim Belk Jun 4 '11 at 23:50
    
I suggest the question being transferred here: quant.stackexchange.com –  vonjd Jun 5 '11 at 11:55
    
There is a useful lesson here if you state the values of the two expressions. A slightly harder question, but still useful, is calculating the probability of coming out ahead after 52 weeks. –  Henry Feb 14 at 1:17
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1 Answer

You are correct. Let's take a closer look at why. In particular, you need to be careful with (b) because it is not true in general for a random variable $X$ that $E[X]$ is the most probable value of $X$. However, it turns out to be true in this case.

For a), If $V$ is the value of the investment after 52 weeks, $$E[V] = \sum_{x=0}^{52} (10000) (1.8)^x (0.4)^{52-x} \binom{52}{x} (0.5)^x (0.5)^{52-x}$$ $$= 10000 \sum_{x=0}^{52} \binom{52}{x} (0.9)^x (0.2)^{52-x} = 10000 (1.1)^{52},$$ which is what you have (and where the last step follows from the binomial theorem).

For b), we can rewrite $V$ as $V = 10000 (0.4)^{52} (1.8/0.4)^X = 10000 (0.4)^{52} (4.5)^X$. Thus $V$ is a one-to-one (and increasing) function of $X$. (This should also be intuitive.) So the most likely value of $V$ occurs when $X$ takes on its most probable value. If $X$ is binomial$(n,p)$, then the most probable value of $X$ (the mode) is $\lfloor (n+1)p \rfloor$. Thus the most likely value of $X$ is $\lfloor 53(0.5) \rfloor = 26$, and so the most likely value of $V$ occurs when $X = 26$, just as you have.

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Thanks for the explanation, it's much clearer now! –  JP83 Jun 8 '11 at 3:43
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