# Finding $\text{Cov}(X,Y)$ when $(X,Y)$ has joint density $\frac{1}{2}\sin(x+y)\mathbf1_{0\le x,y\le\pi/2}$

Joint probability density:

$$$$P_{x,y}(x,y) \begin{cases} \frac{1}{2}\sin(x + y) & , \text{if}\ 0\leq x \leq \frac{\pi}{2}, 0 \leq y \leq \frac{\pi}{2} \\ 0 & , \text{ otherwise} \end{cases}$$$$

What I have done so far:

I know that the equation for covariance is $$E(XY) - E(X)E(Y)$$.

I know that to find $$E(XY)$$ you need to take the double integral $$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\sin(x + y)\,dx\,dy$$.

I'm just not sure how to find $$E(X)$$ and $$E(Y)$$, if they depend on each other

• This is question is related to the marginal probability, you should be able to find the density function for $x$, $y$ by integrating the coupled density with respect to $y$, $x$ -- respectively. – AbuSaad Mar 5 at 7:43

Covariance is $$\int_0^{\pi/2} \int_0^{\pi/2} xy \frac 1 2 \sin(x+y) \, dx \, dy -(\int_0^{\pi/2} \int_0^{\pi/2} x \frac 1 2 \sin(x+y) \, dx \, dy) ( \int_0^{\pi/2} \int_0^{\pi/2} y \frac 1 2 \sin(x+y) \, dx \, dy )$$. I will let you do the computation.