I'm working on a homework problem involving conditional probability.

A simplified model for the movement of the price of a stock supposes that on each day the stock’s price either moves up 1 unit with probability p or moves down 1 unit with probability 1 − p. The changes on different days are assumed to be independent.

What is the probability that after 3 days the stock’s price will have increased by 1 unit?

I believe there are 8 possible outcomes for 3 days, and that 3 of these outcomes would result in the desired change of +1. I think the probability would be 3/8 if both outcomes were equally likely but it doesn't seem that they are.

How do I go about factoring in the p and p-1 information? Is this a binomial problem where the resulting answer would be something like (8 choose 3)(p^3)(p-1)^8? Any hints or helpful resources would be appreciated.

  • $\begingroup$ You're right that this is the binomial distribution, but some of your details seem wrong. There are three "trials", with up being a "success" and down being a "failure". The question is what is the probability of exactly 2 "successes". $\endgroup$ – Ami Tavory Jul 4 '17 at 18:55

You're right there are 8 possible out comes, as each day we have 2 outcomes, and over a total of three days this gives $2^3=8$ outcomes.

The probability of having a total rise of $+1$ after three days is the same as the probability of having a total of $2$ "rises" and $1$ "dips" over these three days, since this will give a total change of $\Delta P = 2 -1 = 1$

So if we let $X$ be the total number of rises out of our three days, it is immediately clear that $X$ is indeed a Binomial random variable, with total number of trials $n=3$ and probability of success $p$ (since the price goes up with probability $p$).

Hence, the probability of having a price change of $+1$ after $3$ days is given by:

$P(X=2) = $$3 \choose 2$$ p^2 (1-p)^1$

Note that this is not identical to the formula you posted in the OP. You've made errors substituting in the values to the Binomial pmf. Here, we substitute in $n=3$,$x=2$ into the formula:

$P(X=x) = $$n\choose x$$p^x (1-p)^{n-x}$

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