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 = -\frac13 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}`