Law of Iterated Expectations and Law of Total Variance

Question

If X is an exponentially distributed random variable, denoted as X ∼ Exp (2).

And E[Y|X] = 15 − 5X + 0.5X^2

Var[Y|X] = 3 + 0.5X

What would be E[Y] and Var[Y] separating the “between” and the “within” components?

For the first part (E[Y]), I got the answer attached, but I don't get a closed-form solution rather than a function. So I'm not sure if it's right or not.

Answer 1

There's a very easy way to proceed...

1. Observe that the simple moments of your $Exp(2)$ rv are the following

$$\mathbb{E}[X^n]=\int_0^{\infty} x^n 2 e^{-2x}dx=\frac{1}{2^n}\int_0^{\infty}(2x)^n e^{-2x}d(2x)=\frac{n!}{2^n}$$

This result has been obtained using Gamma Function

2. Let's calculate $\mathbb{E}[Y]$

$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]=\mathbb{E}\Bigg[15-5X+\frac{X^2}{2}\Bigg]=15-\frac{5}{2}+\frac{1}{4}=\frac{51}{4}$$

3.

$$\mathbb{V}[Y]=\mathbb{E}[\mathbb{V}(Y|X)]+\mathbb{V}[\mathbb{E}(Y|X)]=$$

$$=\mathbb{E}\Bigg[3+\frac{X}{2}\Bigg]+\mathbb{V}\Bigg[15-5X+\frac{X^2}{2}\Bigg]=3+\frac{1}{4}+25\mathbb{V}[X]+\frac{1}{4}\mathbb{V}[X^2]-5Cov(X,X^2)$$

...being

$$\mathbb{V}[X^2]=\mathbb{E}[X^4]-(\mathbb{E}[X^2])^2$$

$$Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$$

The result is

$$\mathbb{V}[Y]=3+\frac{1}{4}+\frac{25}{4}+\frac{1}{4}\Bigg[\frac{4!}{16}-\frac{1}{4}\Bigg]-5\Bigg[\frac{3!}{2^3}-\frac{1}{2}\cdot\frac{1}{2}\Bigg]=\frac{117}{16}$$