# Circle-shaped integration of a bivariate normal distribution

I have a bivariate Normal Distribution of positions in a 2D space given by the marginal distributions of x and y components: $N_x(\mu_x,\sigma_x)$, $N_y(\mu_y,\sigma_y)$.

I want to determine the probability of a coordinate $(x,y)$ to be located within a certain region of the space. In a first step, this region is a rectangle $R$ defined by a tuple of coordinates $R = (x_a, y_a, x_b, y_b)$, which represents two corners.

The probability of $(x,y)$ to be located within this rectangular is

\begin{align} P((x,y) \in R) & = P(x \in [x_a,x_b]) \cdot P(y \in [y_a, y_b]) \\ & = \int_{x_a}^{x_b}p_x(u) du \cdot \int_{y_a}^{y_b}p_y(v)dv \end{align}

The next step is to have a circular region $C$ given by its center $(x_c,y_c)$ and radius $r$.

I know that I have to somehow incorporate the condition $(x-x_c)^2 + (y-y_c)^2 = r^2$ into the integration, which introduces a non-trivial dependency between the two dimensions. And at this point I am lost.

So I want to ask you, how to best calculate $P((x,y) \in C)$?

EDIT

Thanks to Henry for his inspiring comment! It's now obvious to me, how to solve this. However, integration boundaries are dependent and non-constant. This makes it difficult for me, to use this solution in my special case.

Is the following transformation into polar coordinates equivalent? $$\int_{0}^{r} \int_{0}^{2\pi} p_x(x_c + t\cdot cos(\phi)) \cdot p_y(y_c + t\cdot sin(\phi)) d\phi dt$$

• Presumably the two random variables are independent – Henry Jul 17 '18 at 7:54
• It may not be the best way, but I think one way of writing this is $$\int_{u=-r}^{r} p_x(x_c+u) \int_{v=-\sqrt{r^2-u^2}}^{\sqrt{r^2-u^2}} p_y(y_c+v)\,dv \, du$$ – Henry Jul 17 '18 at 8:00