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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 $.

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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 $P(A)=P(B)=0.5$

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  • $\begingroup$ So I got $3[(0.7)^2(0.3)]$ for the probability that stock B goes down by 1 unit after 3 days. This would correspond to P(Down 1 unit|B)? $\endgroup$ – gofish Jan 18 '17 at 3:35
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Momo Jan 18 '17 at 3:39
  • $\begingroup$ @Momo for b) shouldn't the probability be 1/8? Like in a, the stock can only be down 1 after 3 days, after 2 down and 1 up day; this happens in 3!/2! = 3 ways. However, of these outcomes, only 1 involves the stock rising on the first day (UDD, DUD, DDU); since P(down 1 after 3) = P(up 1 after 3) = 3/8 then P(down after 3|up first day) = 3/8*1/3 = 1/8 $\endgroup$ – M.Diggerson Jan 18 '17 at 3:42
  • $\begingroup$ (b) is asking for P(Day 1 = U | up 1 after 3 days). So in the space of possible results that are up 1 after 3 days, there are 3 ways it can happen and only 1 starts with U, hence 1/3. $\endgroup$ – ConMan Jan 18 '17 at 3:45
  • $\begingroup$ I should have probably also said that $$P(\text{Down on the first day}|\text{Up 1 unit})=\frac{P(\text{Down on the first day and Up 1 unit})}{P(\text{Up 1 unit})}=\frac{P(\{\text{DUU}\})}{P(\{\text{DUU},\text{UDU},\text{UUD}\})}=\frac{1/8}{3/8}=\frac{1}{3}$$ $\endgroup$ – Momo Jan 18 '17 at 3:55

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