Expectation calculation of a function for bivariate normal distribution

Question

I am struggling with parts b and c of the question below, covariance value is 1/2, for the remaining parts, any help would be much appreciated

Answer 1

a) ok

b) the condition $\mathbb{E}(X+Y)=0$ implies also $V(X+Y)=\mathbb{E}(X+Y)^2$

to finish the proof it is enough observe that

$\mathbb{E}|X+Y|=\mathbb{E}\sqrt{(X+Y)^2}$

and using Jensen's Inequality you immediately get:

$$\mathbb{E}[\sqrt{(X+Y)^2}] \leq \sqrt{\mathbb{E}(X+Y)^2}$$

c) it is very easy,

$(X+Y)\sim N(0;3)$ thus $f_{|X+Y|}(t)=\frac{2}{\sqrt{6\pi}}e^{-\frac{t^2}{6}}$; $t\geq 0$

Now,

$$\mathbb{E}|X+Y|^3=\int_0^{+\infty}t^3\frac{2}{\sqrt{6\pi}}e^{-\frac{t^2}{6}}dt$$

Letting $\frac{t^2}{6}=u$ the integral becomes

$$6\sqrt{\frac{6}{\pi}}\int_o^{+\infty} ue^{-u}du=6\sqrt{\frac{6}{\pi}}$$

Anyway when posting a question show your efforts (using mathjax an not pictures)