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 $

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  • $\begingroup$ 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 $\begin{pmatrix}\mu_1\ \\mu_2\end{pmatrix}$ $\endgroup$ – BlockchainDieter Nov 22 '18 at 8:58
  • $\begingroup$ $g(x_1,x_2)$ 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 $\mu_2$, $g(x_1,x_2) = x_2$. $\endgroup$ – user158565 Nov 22 '18 at 14:33

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