PDF of projection of uniform distributed random variable

Question

With the usual p-norm, let $B_1^p(0):= \{ x\in \mathbb{R}^2 : |x|_p \leq 1 \}$ for a $p\in[1,\infty]$. Let $X_p$ be a uniformly distributed random variable on $B_1^p(0)$.

Calculate the PDF of the random variable $\pi_1(X_p)$ where $\pi_1:\mathbb{R}^2 \to \mathbb{R}$ is the projection on the first coordinate.

So thus far I have the following idea. The density should be $$f_{X_p}(x_1,x_2) = \frac{1_{B_1^p(0)}(x_1,x_2)}{\lambda_2({B_1^p(0)})} $$

Could the corresponding density $f_{\pi_1(X_P)}$ be just the Integral of the first density but in respect to $dx_2$ and maybe more generally, is this the right approach to define the uniform distribution on any set?

Answer 1

Let $X=(X_1,X_2)$. As suggested in the comments, the answer is $$\frac{d}{dz}P(X_1\leq z)=\frac{d}{dz}\frac{2}{\lambda^2(B^p(0))}\int_{-1}^z(1-|x|^p)^{1/p}dx=\frac{2(1-|z|^p)^{1/p}}{\lambda^2(B^p(0))},z \in [-1,1]$$ where $\lambda^2(B^p(0))=\frac{(2\Gamma(1+1/p))^2}{\Gamma(1+2/p)}$.