# Solved (Free): As a measure of intelligence, mice are timed when going through a maze to reach a reward of food. The time (in seconds) required for any mouse

#### ByDr. Raju Chaudhari

Apr 6, 2021

As a measure of intelligence, mice are timed when going through a maze to reach a reward of food. The time (in seconds) required for any mouse is a random variable Y with a density function given by

 $$\begin{equation*} f(y)= \begin{cases} \frac{b}{y^2} & y\geq b\\ 0 & \text{ otherwise }. \end{cases} \end{equation*}$$

a. Show that $f(y)$ has the properties of a density function.
b. Find $F(y)$.
c. Find $P(Y > b + c)$ for a positive constant $c$.
d. If $c$ and $d$ are both positive constants such that $d > c$, find $P(Y > b + d|Y > b + c)$.

#### Solution

a.

(i) $f(y)\geq 0$ for all $y$
(ii)

 \begin{aligned} \int_b^\infty f(y)\; dy &= \int_b^\infty \frac{b}{y^2}\; dy\\ &=\bigg[-\frac{b}{y}\bigg]_b^\infty\\ &= -(0-1)\\ &= 1 \end{aligned}

Hence $f(y)$ is a density function.

b. Distribution function of $Y$ is

 \begin{aligned} F(y) &= P(Y\leq y)\\ &= \int_b^y f(t)\; dy \\ &= \int_b^y \frac{b}{t^2}\; dy\\ &= \bigg[-\frac{b}{t}\bigg]_b^y\\ &= \bigg[-\frac{b}{y}+\frac{b}{b}\bigg]\\ &= 1-\frac{b}{y} \end{aligned}

c. $P(Y> b+c)$ is

 \begin{aligned} P(Y > b+c) &= 1- P(Y\leq b+c)\\ &= 1- F(b+c)\\ &= 1-\bigg(1-\frac{b}{b+c}\bigg)\\ &= \frac{b}{b+c} \end{aligned}

d. $P(Y > b+d|Y > b+c)$ is

 \begin{aligned} P(Y> b+d|Y> b+c) &= \frac{P[(Y>b+d)\cap (Y> b+c)]}{P(Y> b+c)}\\ &=\frac{P(Y> b+d)}{ P(Y\leq b+c)} \end{aligned}

$Y>b+d$ and $Y> b+c$ if $Y>b+d$ (because d>c).

 \begin{aligned} P(Y > b+d|Y > b+c) &=\frac{P(Y > b+d)}{ P(Y\leq b+c)}\\ &= \frac{1- P(Y < b+d)}{P(Y > b+c)}\\ &= \frac{1- P(Y < b+d)}{1-P(Y < b+c)}\\ &= \frac{1- \bigg(1-\frac{b}{b+d}\bigg)}{1- \bigg(1-\frac{b}{b+c}\bigg)}\\ &= \frac{\frac{b}{b+d}}{\frac{b}{b+c}}\\ &=\frac{b+c}{b+d} \end{aligned}