Question

g A random variable X has a probability density function fX(x) = ? 0.5 sin(x) , 0 ≤ x ≤ π 0 , otherwise Another random variable Y is defined as Y = X + (X − π/2)3 . (a) What are the mean (expected value), variance and standard deviation of X? (b) Plot the function Y = Y (X). What are the minimum and maximum possible values of Y ? (c) From the plot, using only graphical thinking and no equations, what is the shape of the probability density function of Y (i.e., fY (y))? Specifically, how is the shape compared to that of fX(x)? (d) From the plot, using only graphical thinking and no equations, estimate (guess) the expected value and standard deviation of Y (i.e., µY and σY ).

Answer 1

$f_X(x)=\begin{cases}0.5\sin x&\text{for }0\le x\le\pi\\0&\text{otherwise}\end{cases}$

a. The mean of $X$ is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac12\int_0^\pi x\sin x\,\mathrm dx=\frac\pi2$

Recall that the variance of a random variable $X$ is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

We have

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac12\int_0^\pi x^2\sin x\,\mathrm dx=\frac{\pi^2-4}2$

so that

$\mathrm{Var}[X]=\dfrac{\pi^2-4}2-\dfrac{\pi^2}4=\dfrac{\pi^2-8}4$

and the standard deviation is

$\sqrt{\mathrm{Var}[X]}=\dfrac{\sqrt{\pi^2-8}}2$

b.

$X=0\implies Y=-\dfrac{\pi^3}8$

$X=\pi\implies Y=\pi+\dfrac{\pi^3}8$