# Let $X$ be normal with zero mean and variance $\sigma^2$, let $Y$ be uniform on $(0,\pi)$. Find the density of $Z=X+a \cos(Y)$.

Let $X$ be normal with zero mean and variance $\sigma^2$, let $Y$ be uniform on $(0,\pi)$ and let $a$ be a real number. Assume $X$ and $Y$ are independent . Find the density of $Z=X+a \cos(Y)$.

I just need to find $P(Z \leq t)=\int_0^\pi \int_{-\infty}^{a \cos(Y)+t} f_X (x) \frac{1}{\pi} \, dx \, dy$ and take the derivative with respect to $t$?

## 1 Answer

The density of the sum of independent random variables is the convolution of their two densities. The random point $a(\cos Y, \sin Y)$ is uniformly distributed on the circle. The length of the arc of a circle whose points have $y$-coordinates in any set $A$ on the $y$-axis is $$\int_A a\,d\theta = \int_A \frac{a\,dy}{\sqrt{a^2-y^2}}.$$ Find the normalizing constant $c$ that will make $$\int_{-a}^a \frac{c\,dy}{\sqrt{a^2-y^2}} =1$$ and you'll have the density function of the random variable $Y$.

The convolution is $$f(z) = \int_{-a}^a \frac{c}{\sqrt{a^2-y^2}} \cdot \frac 1 {\sqrt{2\pi}} e^{-(z-y)^2/2} \,dy.$$ I'm inclined to doubt there is a closed form for this.

• PS: Above I assumed $\sigma=1$. For other values of $\sigma$, mutatis mutandis.... – Michael Hardy Jul 29 '16 at 19:11