Question

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.

(a) What is the probability that after$2$days the stock will be at its original price?

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

(c) Given that after $3$days the stock’s price has increased by $1$ unit, what is the probability that it went up on the first day?

Short answer

a) The probability that after 2 days the stock will be at its original price is$2p(1-p)$

b) The probability that after 3 days the stock’s price will have increased by 1 unit is$3p^2(1-p)$

c) The probability that it went up on the first day is $\frac{2}{3}$

Step-by-step solution

Given information (part a)

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

Explanation(part a)

Name:

$U_i$- event that the stock went up in the$i-th$day

$D_i=U_i^c$- event that the stock went down in the $i-th$day

Probabilities:

$P\left(U_i\right)=p\Rightarrow P\left(D_i\right)=1-p$

The event whose probability is requested is a union of two events:

$U_1{D}_2\cup D_1{U}_2$

The events are mutually exclusive therefore

the probability is:

$P\left(U_1{D}_2\cup D_1{U}_2\right)=P\left(U_1{D}_2\right)+P\left(D_1{U}_2\right)$

$=P\left(U_1\right)P\left(D_2\right)+P\left(D_1\right)P\left(U_2\right)$independence of events $U_1$and $U_2$

$=p(1-p)+(1-p)p$

$=2p(1-p)$

Final answer(part a)

The events are mutually exclusive therefore the probability is:

$=2p(1-p)$

Step 4:Given information(part b)

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$1-p$

Step 5:Explanation(part b)

If the stock's price has increased by $1$unit after $3$days, it will move down $1$unit in one of three days and move up $2$units in the remaining two days.

Hence, the probability that after $3$days the stock's price will have increased by $1$unit is$3p^2(1-p)$

Final answer (part b)

The events are mutually exclusive therefore the probability is:

$=3p^2(1-p)$

Step 7:Given information(part c)

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$1-p$

Explanation(part c)

Name:

$U_i$- event that the stock went up in the $i-th$day

$D_i=U_i^c-$event that the stock went down in the$i-th$day

Probabilities

$P\left(U_i\right)=p\Rightarrow P\left(D_i\right)=1-p$

Use the definition of conditional probability

$P\left(U_1\mid U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)=\frac{P\left(U_1\cap \left(U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)\right)}{P\left(U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)}$

For the numerator

$U_1\cap \left(U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)=U_1{U}_2{D}_3\cup U_1{D}_2{U}_3$

$\Rightarrow P\left(U_1{U}_2{D}_2\cup U_1{D}_2{U}_3\right)=P\left(U_1{U}_2{D}_3\right)+P\left(U_1{D}_2{U}_3\right)$

$=P\left(U_1\right)P\left(U_2\right)P\left(D_3\right)+P\left(U_1\right)P\left(D_2\right)P\left(U_3\right)$

$=2p^2(1-p)$

Divide the numerator with the result of part (b)

ie,

$P\left(U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)=$$3p^2(1-p)$

We get,

$P\left(U_1\mid U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)=\frac{2}{3}$

Final Answer(part c)

Probability to went up on on stock price the first day$P\left(U_1\mid U_1{D}_2{U}_3\cup D_1{U}_2{U}_3\cup U_1{U}_2{D}_3\right)=\frac{2}{3}$