# Mean and variance of probability density with multidimensional indicator function

I encountered the following question while studying machine learning:

We are asked to calculate mean and covariance of a given probability density function

$$p(x) = \frac{1}{16} \cdot 1_{0 \leq x_1 \leq 4} \cdot 1_{-1 \leq x_2 \leq 3}$$

where $$x \in \mathbb{R}^2$$ and $$1$$ is an indicator function.

I know the mean in a one-dimensional setting can be computed like so:

$$\mu = \int p(x) x dx$$

What confuses me here are two things:

1. How to compute the mean in a two-dimensional setting?
2. If it involves integration: How to integrate the indicator functions?

For the two dimensional random vector $$(X_1,X_2)$$ with pdf $$f(x_1,x_2)$$, you have two ways to get the expectations involved one random variables, for $$g(X_1)$$. 1) you get the marginal pdf of $$X_1$$ first, then following the way that you get thge expectation for one random variables. 2) $$\int \int g(x_1)f(x_1,x_2)dx_1 dx_2$$. For the expectation involved two random variables, $$g(x_1,x_2)$$, for example, $$E(X_1X_2)$$, you must go through $$\int \int g(x_1,x_2)f(x_1,x_2)dx_1 dx_2$$.
For question 2. Here is example: to show $$\int\int p(x_1,x_2) d_x1 dx_2 =1$$, so that it meet one of the requirements to be pdf. For the area that pdf = 0, the integral is zero on the, so you just to consider the area that pdf not zero. In you case, the area is $$0 \leq x_1 \leq 4$$ and $$-1 \leq x_2 \leq 3$$.
$$\int_{-1}^{3}\int_{0}^{4} p(x_1,x_2) dx_1 dx_2 = \int_{-1}^{3}\int_{0}^{4} \frac 1{16} dx_1 dx_2 = \int_{-1}^{3}\frac {x_1}{16}|_0^4 dx_2 = \int_{-1}^{3}\frac {1}{4} dx_2 = \frac {x_2}{4}|_{-1}^3 = \frac 34 - (-\frac 14) = 1$$
• Thanks! So if I understand correctly, in this specific case $g(x_1, x_2) = \begin{pmatrix}x_1\\x_2\end{pmatrix}$? And would it be sufficient to calculate $\int \int g(x_1)f(x_1,x_2)dx_1 dx_2$ for $x_1$, $x_2$ respectively and then outputting the mean as $\begin{pmatrix}\mu_1\ \\mu_2\end{pmatrix}$ – BlockchainDieter Nov 22 '18 at 8:58
• $g(x_1,x_2)$ should be single value function. I do not know how to work with $g(x_1, x_2) = \begin{pmatrix}x_1\\x_2\end{pmatrix}$. For $\mu_1$, $g(x_1,x_2) = x_1$, For $\mu_2$, $g(x_1,x_2) = x_2$. – user158565 Nov 22 '18 at 14:33