Question

The pdf of a random variable x is given by

f(x)={3/2x²,-1<y<1 both inclusive

{0,elsewhere

Find the pdf of y=(1/2)x². Hence state the mean and standard deviation of y

Answer 1

The probability distribution function (PDF) of the given random variable X is

$f_X(x)=\frac{3}{2}x^2$, for $-1\leq x\leq 1$ and $f_X(x)=0$ elsewhere.

The cumulative distribution function (CDF) of X is

$F_X(x)=\int\limits_{-\infty}^{x}f_X(t)dt$

Hence

$F_X(x)=0$, for x<1

$F_X(x)=1$, for x>1

$F_X(x)=\int\limits_{-1}^x\frac{3}{2}t^2dt=\frac{1}{2}t^3|_{-1}^x=(x^3+1)/2$ for $-1\leq x\leq 1$.

Consider the random variable $Y=X^2/2$.

$F_Y(y)=P(Y\leq y)=P(X^2/2\leq y)=P(X^2\leq 2y)$

Hence

$F_Y(y)=0$, if y<0

$F_Y(y)=1$, for 2y>1 (i.e. y>1/2)

$F_Y(y)=F_X(\sqrt{2y})-F_X(-\sqrt{2y})=$

$(1+(\sqrt{2y})^3)/2-(1-(\sqrt{2y})^3)/2=2\sqrt{2}y^{3/2}$

Then

$f_Y(y)=\frac{d}{dy}F_Y(y)$

$f_Y(y)=3\sqrt{2}y^{1/2}$ for $y\in[0,1/2]$ and $f_Y(y)=0$ elsewhere.

Expectation of $Y$ is $E(Y)=\int\limits_0^{1/2} yf_Y(y)dy=\int\limits_0^{1/2}3\sqrt{2}y^{3/2}dy=$

$=3\sqrt{2}\cdot \frac{2}{5}y^{5/2}|_0^{1/2}=3/10=0.3$

Expectation of $Y^2$ is $E(Y^2)=\int\limits_0^{1/2} y^2f_Y(y)dy=\int\limits_0^{1/2}3\sqrt{2}y^{5/2}dy=$

$=3\sqrt{2}\cdot \frac{2}{7}y^{7/2}|_0^{1/2}=3/28$

Variance of $Y$ is $V(Y)=E(Y^2)-E(Y)^2=3/28-9/100=3/175$

Standard deviation is $\sigma_Y=\sqrt{V(Y)}=\sqrt{3/175}=\sqrt{21}/35=0.131$

Answer. $E(Y)=0.3$; $\sigma_Y=0.131$.