I'm trying to understand something about marginal distributions. Let $X:\Omega \rightarrow \mathbb{R}^n$ be a random variable with density $\alpha$ and $v \in \mathbb{R}^n$ a unit vector. Now consider the real random variable $Y:= \langle v, X \rangle$. What ist the density of $Y$?
This is my idea: If we denote $V:= \mathrm{span}(v)$ and $W$ as the orthogonal complement of $V$ then $V \cong \mathbb{R}, W \cong \mathbb{R}^{n-1}$, we can write $\mathbb{R}^n = V \times W$ and $\alpha(x) = \alpha(u,w)$ with $u \in V$ and $ w \in W$. Let $f_Y(y)$ denote the density function of $Y$. Intuitively I would say that $f_Y(y) = \int_W \alpha(u,w) dw$. Is it correct and if yes, how can I prove it? Also, is it relevant that $v$ is a unit vector?
Thank you!
You have the right idea but you have to be careful about the way you do the change of variable from $\mathbb{R}^n$ to $V\times W$.
One way to show that a random variable $Y$ of $\mathbb{R}$ has a density of $f_Y(y)$ is to show that for any Borel function $\phi:\mathbb{R}\rightarrow\mathbb{R}$ then $$\mathbb{E}[\phi(Y)]=\int_{\mathbb{R}}\phi(y)f_Y(y)\mathrm{d} y.$$ In your case, we have $$\mathbb{E}[\phi(Y)]=\mathbb{E}[\phi(\langle v,X\rangle)]=\int_{\mathbb{R}^d}\phi(\langle v,x\rangle)\alpha(x)\mathrm{d} x$$ Now, every $x\in\mathbb{R}^n$ can be uniquely written as $y\times v + w$ where $y\in\mathbb{R}$ and $w\in W$ since $W$ is the orthogonal complement of $V$. We can then do the change of variable $x\mapsto (y,w)$ in our integral and we get $$\mathbb{E}[\phi(Y)]=\int_\mathbb{R}\int_W \phi(\langle v,y\times v + w\rangle)\alpha(y\times v+w)\mathrm{d} w \mathrm{d} y=\int_\mathbb{R} \phi(y)\left(\int_W\alpha(y\times v+w)\mathrm{d} w\right)\mathrm{d} y$$ and we can finally conclude that $$f_Y(y)=\int_W\alpha(y\times v+w)\mathrm{d} w.$$ Note that the absolute value of the determinant of the Jacobian matrix for the change of variable $x\mapsto (y,w)$ is equal to $1$ because $v$ is a unit vector so that's one place where this assumption was important. Another place in the proof where the assumption that $v$ is a unit vector was usefull is when we used the identity $$\langle v,y\times v + w\rangle = y\times \langle v,v\rangle + \langle v,w\rangle = y.$$