Expected value and variance of a random vector

I am trying to solve expected value and variance of the random vector given by function: $$$$p_x(x)=\left\{ \begin{array}{@{}ll@{}} \frac1\pi, & \text{if}\ x^2_1+x^2_2 < 1 \\ 0, & \text{otherwise} \end{array}\right.$$$$ Condition $$x^2_1+x^2_2$$ let me know that this is an unit circle in 3d-plane, so I can integrate from [-1 1] in both directions: $$E(x)=\begin{bmatrix} \int^1_{-1}{x_1*\frac1\pi} \\ \int^1_{-1}{x_2*\frac1\pi} \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ Is that right way of doing expected value? What about variance? I didnt quite catche how to figure out the variance. I have now probability theory course and some other course, which uses that part of probability theory. The other course is going on and probability theory has just started. Because of that I have asked questions here. Thanx.

• For variance, you need to evaluate $\int\int x^2_1 p(x_1,x_2)dx_1dx_2$. This should be straightforward by integrating $x_2$ from $-\sqrt{1-x_1^2}$ to $\sqrt{1-x_1^2}$ and $x_1$ from $-1$ to $1$. – Alex R. Jan 10 at 17:52
• Is thar right way then: $$\int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}\int_{-1}^1 x_1^2 \frac1\pi dx_1 dx_2$$ ? – Hillbilly Joe Jan 10 at 18:03

It is often the case that some introductory course on probability theory requires some knowledge from multivariate calculus. To evaluate the variance matrix, it is much more efficient to use polar coordinate change $$(x_1,x_2)=(r\cos\theta,r\sin \theta).$$ Then $$dx_1dx_2 = rdrd\theta$$, and we have $$\frac{1}{\pi}\int_{x_1^2+x_2^2<1}x_1^2\;dx_1dx_2=\frac{1}{\pi}\int_0^{2\pi}\int_0^1 r^3\cos^2 \theta\;drd\theta=\frac{1}{4},$$ $$\frac{1}{\pi}\int_{x_1^2+x_2^2<1}x_1x_2\;dx_1dx_2=\frac{1}{\pi}\int_0^{2\pi}\int_0^1 r^2\cos\theta\sin\theta\;drd\theta=0,$$ and the variance of $$X_2$$ is the same as that of $$X_1$$.