Expected value of a inner product of vectors

Question

One has a vector of random variables $$x=[{x_1,x_2}]$$ that each is i.i.d as $$N(0,2)$$ So, $$Var[x_i]=E[x_i^2]-E[x_i]^2=>E[x_i^2]=2$$for $i=1,2.$Need to find the distribution of $$x^Tx$$ Hence, $$E[x^Tx]=E[x_1^2+x_2^2]=E[x_1^2]+E[x_2^2]=4$$ Next,$$Var[x^Tx]=E[(x^Txx^Tx)]-E[x^Tx]^2=E[(x_1^2+x_2^2)(x_1^2+x_2^2)]-E[x_1^2+x_2^2]^2=E[x_1^4+x_2^4+2x_1^2x_2^2]-(E[x_1^2]+E[x_2^2])^2=E[x_1^4]+E[x_2^4]+2E[x_1^2]E[x_2^2]-(4)^2=E[x_1^4]+E[x_2^4]+2*2*2-16=E[x_1^4]+E[x_2^4]-8$$ Is there any way to get a numerical value for the variance? Maybe I am doing something wrong.

Answer 1

You want $\operatorname{Var}[x_1^2 + x_2^2]$. Since $x_1$, $x_2$ are iid, so are $x_1^2$ and $x_2^2$; thus the variance of the sum is the sum of the variances. Your difficulty is computing the higher moment $\operatorname{E}[x_i^4]$. But since $x_i$ is normal with mean $0$ and variance $2$, we can either compute it mechanically via integration; e.g. $$\operatorname{E}[x_i^4] = \int_{x=-\infty}^\infty x^4 \cdot \frac{1}{2\sqrt{\pi}} e^{-x^2/4} \, dx$$ which is an exercise in integration by parts; or you can appeal to the properties of the moment-generating function of a normal distribution with mean $\mu$ and variance $\sigma^2$: $$M_X(t) = \operatorname{E}[e^{tX}] = e^{\mu t + (\sigma t)^2/2},$$ for which you have the special case $$M_{x_i}(t) = e^{t^2},$$ and $$\operatorname{E}[x_i^4] = \frac{d^4}{dt^4} \left[ M_{x_i}(t) \right]_{t=0} = \left[4e^{t^2}(3 + 12t^2 + 4t^4)\right]_{t=0} = 12.$$ The rest is left as a straightforward computational exercise. The third method is to recognize that the variable $$Z = \frac{x_i}{\sqrt{2}} \sim \operatorname{Normal}(0,1)$$ is standard normal, thus $$Z^2 = \frac{x_i^2}{2} \sim \operatorname{ChiSquare}(\nu = 1)$$ and $$\frac{x_1^2 + x_2^2}{2} \sim \operatorname{ChiSquare}(\nu = 2).$$ Thus this variable has variance of $2\nu = 4$, i.e. $$\operatorname{Var}\left[\frac{x_1^2 + x_2^2}{2}\right] = 4,$$ from which you can easily determine the desired variance. All three methods are legitimate and yield the same result.

Answer 2

The inner product will be a Chi squared random variable with $k=2$ degrees of freedom. Thus the variance is $2\times k\times2^2=16$.