Question

Consider the cumulative distribution function (cdf)F(y) =0, y≤0,y2,0< y≤1,11≤y.

(a) Find E(Y) ifYhas this cdf.

(b) Find P(Y >1/3|Y≤2/3) ifYhas this cdf.

(c) SupposeY1andY2are a random sample from this distribution. FindP(Y1≤1/2,Y2>1/2).

Answer 1

The cumulative distribution function of Y is:

For $y\leq 0$ , $F(y)=0$

For $0<y\leq 1$ , $F(y)=y^{2}$

For , $F(y)=1$

(a)

The density function of Y can be obtained by differentiating the distribution function with respect to y as:

For $0<y\leq 1$ , $f(y)=2y$

otherwise, $f(y)=0$ .

Hence, the E(Y) can be obtained as:

$E(Y)=\int_{0}^{1}y\cdot f(y)dy$

$E(Y)=\int_{0}^{1}y\cdot (2y)dy$

$E(Y)=\left [ \frac{2y^{3}}{3} \right ]_{y=0}^{y=1}$

$E(Y)=\left [ \frac{2(1)^{3}}{3}-\frac{2(0)^{3}}{3} \right ]$

$E(Y)=0.6667$

(b)

The probability of the following expression can be obtained as:

$P(Y>\frac{1}{3}|Y\leq \frac{2}{3})=\frac{P(\frac{1}{3}< Y\leq \frac{2}{3})}{P(Y\leq \frac{2}{3})}$

$P(Y>\frac{1}{3}|Y\leq \frac{2}{3})=\frac{P(Y<\frac{2}{3})-P(Y<\frac{1}{3})}{P(Y< \frac{2}{3})}$

$P(Y>\frac{1}{3}|Y\leq \frac{2}{3})=\frac{F(\frac{2}{3})-F(\frac{1}{3})}{F(\frac{2}{3})}$

$P(Y>\frac{1}{3}|Y\leq \frac{2}{3})=\frac{(\frac{2}{3})^{2}-(\frac{1}{3})^{2}}{(\frac{2}{3})^{2}}$

$P(Y>\frac{1}{3}|Y\leq \frac{2}{3})=0.75$

(c)

Suppose Y1 and Y2 are a random sample from this distribution.

Assume Y1 and Y2 are independent and identically distributed.

The probability of the following expression can be obtained as:

As the variables are independent, hence,

$P(Y_{1}\leq\frac{1}{2},Y_{2}> \frac{1}{2})=P(Y_{1}\leq\frac{1}{2})\times P(Y_{2}> \frac{1}{2})$

$P(Y_{1}\leq\frac{1}{2},Y_{2}> \frac{1}{2})=F(\frac{1}{2})\times (1-P(Y_{2}< \frac{1}{2}))$

$P(Y_{1}\leq\frac{1}{2},Y_{2}> \frac{1}{2})=F(\frac{1}{2})\times (1-F(\frac{1}{2}))$

As the variables are identical, hence,

$P(Y_{1}\leq\frac{1}{2},Y_{2}> \frac{1}{2})=(\frac{1}{2})^{2}\times (1-(\frac{1}{2})^{2})$

$P(Y_{1}\leq\frac{1}{2},Y_{2}> \frac{1}{2})=0.1875$