Anthony owns Stock A. Suppose that on any given day, the price of Stock A will either go up
or down by $1$ unit with equal probability. Also assume that the daily movements in price are
independent over different days.
(a) What is the probability that the stock price has gone up by $1$ unit after $3$ days?
(b) Suppose that after $3$ days, the stock price has gone down by $1$ unit. What is the probability
that the price went up on the first day?
(c) Anthony’s friend, Brandon, owns Stock B. On each day, the price of Stock B will either go
up or down by $1$ unit. The probability that the price will go up on a single day is $0.3$. Again,
the daily movements in price are independent over different days.
One of the two stocks is randomly chosen. Suppose that after $3$ days the price of the chosen
stock has gone down by $1$ unit. What is the probability that the selected stock is Stock A?
So I have done part a). Not sure if it is correct, but this is what I have:
a) $P(\text{going up 1 unit}) = P(\text{going down 1 unit}) = p$
For the stock to go up $1$ unit after $3$ days, then it would be $3p^3
$.
Best Answer
First, if the stock goes either up or down, than $p=\frac{1}{2}$
For a) you need to have UUD UDU or DUU with probability $\frac{1}{8}$ each, so the answer is indeed $\frac{3}{8}$
For b) you need to have UDD DUD or DDU with probability $\frac{1}{8}$ each, so the answer is $\frac{1}{3}$
For c) use Bayes' theorem.
$P(A|\text{Down 1 unit})=\frac{P(\text{Down 1 unit|A})P(A)}{P(\text{Down 1 unit|A})P(A)+P(\text{Down 1 unit|B})P(B)}$
$P(A|\text{Down 1 unit})$ and $P(B|\text{Down 1 unit})$ are the probability of one success in $3$ independent trials (binomial random variable), when the probability of success for each trial is $\frac{1}{2}$ and $0.3$ respectively.
$P(\text{Down 1 unit|A})={3\choose 1}\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^2=\frac{3}{8}=0.375$
$P(\text{Down 1 unit|B})={3\choose 1}0.3^10.7^2=0.441$
and $P(A)=P(B)=0.5$