Question

chegg Y=4X−2 XX has a PDF of f_X(x)=\begin{cases} 3e^{-3x} & 0 \leq x \\ 0 & \text{otherwise}\end{cases}f X ​ (x)={ 3e −3x 0 ​ 0≤x otherwise ​ What is E[Y^2]E[Y 2 ]?

Answer 1

$\Bbb E[Y^2] = \Bbb E[(4X - 2)^2]$

so by definition of expectation,

$\Bbb E[Y^2] = \displaystyle \int_{-\infty}^\infty (4x-2)^2 f_X(x) \, dx = \int_0^\infty 3(4x-2)^2 e^{-3x} \, dx$

Integrate by parts (twice).

$\displaystyle \int_a^b u\,dv = uv\bigg|_a^b - \int_a^b v\,du$

First, let

$u = (4x-2)^2 \implies du = 8(4x-2)\,dx \\\\ dv = 3e^{-3x} \, dx \implies v = -e^{-3x}$

so that

$\displaystyle \Bbb E[Y^2] = -(4x-2)^2 e^{-3x} \bigg|_{x=0}^{x\to\infty} + 8 \int_0^\infty (4x-2) e^{-3x} \, dx \\\\ ~~~~ = 4 + 8 \int_0^\infty (4x-2) e^{-3x} \, dx$

Next,

$u = 4x-2 \implies du = 4\,dx \\\\ dv = e^{-3x} \, dx \implies v = -\dfrac13 e^{-3x}$

so that

$\displaystyle \Bbb E[Y^2] = 4 + 8 \left(-\frac13 (4x-2) e^{-3x} \bigg|_{x=0}^{x\to\infty} + \frac43 \int_0^\infty e^{-3x} \, dx\right) \\\\ ~~~~ = 4 + 8 \left(-\frac23 - \frac49 e^{-3x}\bigg|_{x=0}^{x\to\infty}\right) \\\\ ~~~~ = 4 - \frac{16}3 + \frac{32}9 = \boxed{\frac{20}9}$