8
$\begingroup$

Let $U \subset \mathbb{R}^n$ be a vector space with $\dim(U)=d$. A standard normal distribution on $U$ is the law of a random vector $X=(X_1, \ldots, X_n)$ taking values in $U$ and such that the coordinates of $X$ in one ($\iff$ in any) orthonormal basis of $U$ is a random vector made of $d$ independent standard normal distributions ${\cal N}(0, 1)$.

When reading this question I asked myself the following question. Let $Y=(Y_1, \ldots, Y_n)$ be a standard normal distribution on $\mathbb{R}^n$. Is is true that the conditional distribution of $Y$ given $Y \in U$ is the standard normal distribution on $U$ ?

The squared norm ${\Vert X \Vert}^2$ of $X$ has a chi-square distribution $\chi^2_d$. Thus, if this is true, that would explain @Argha's claim.

Sorry if the LaTeX is mistyped, I don't see the LaTeX rendering :(

EDIT 01/10/2012: Ok I see. Write $y=u+v$ the orthogonal decompostion of $y$ in $U\oplus U^\perp$. Then $$\Pr(Y\in \mathrm{d}y \cap Y \in U)=\Pr(P_U Y \in \mathrm{d}u)$$. That shows that $(Y \mid Y \in U) \sim P_U Y$. This is little bit heuristic but morally correct. Finally it is clear from the definition that $P_U Y$ is standard normal on $U$.

$\endgroup$
4
  • 2
    $\begingroup$ Isn't this terribly obvious when you note that an orthonormal basis for $\mathbb{R}^n$ can always be constructed by extending any orthonormal basis for $U$? (One proof: use Gram-Schmidt on any extension, whether orthonormal or not.) In this basis the PDF is separable and a fortiori is standard normal on $U$, QED. $\endgroup$ – whuber Sep 27 '12 at 14:35
  • $\begingroup$ @whuber Please could you elaborate in an answer ? How do you derive the conditional distribution ? $\endgroup$ – Stéphane Laurent Sep 27 '12 at 14:53
  • 3
    $\begingroup$ You just look at it! When an absolutely continuous PDF $f(x,y)$ factors as $f_x(x)f_y(y)$, then (a) $X$ and $Y$ are independent and (b) $f_x$ and $f_y$ are the conditional distributions. $\endgroup$ – whuber Sep 27 '12 at 15:48
  • $\begingroup$ @whuber I'm just coming back from work. I will think about this later. Thanks. Of course I believe this is obvious but I'm tired. $\endgroup$ – Stéphane Laurent Sep 27 '12 at 16:56
3
$\begingroup$

Yes. You have that $U$ is a subspace of $\mathbb R^n$. Let $Y \sim \text{N}(0,I)$ and $P$ be the orthogonal projection matrix on $U$, so that $P$ is symmetric and idempotent. Then $PY \sim \text{N}(P0,PIP^T) = \text{N}(0,P)$. This is a singular normal distribution, which on the subspace $U$ is the standard normal on that subspace. As a singular distribution, it does not have a density with respect to volume measure in $\mathbb R^n$, but it does have a density with respect to the (lower-dim) volume measure on $U$.

$\endgroup$
9
  • $\begingroup$ I don't see where do you prove that $PY$ has the same law as $Y$ conditional to $Y \in U$ ? $\endgroup$ – Stéphane Laurent Sep 27 '12 at 14:01
  • $\begingroup$ Note that abstractly, conditional probability (really expectation, to get a linear space ...) is a projection! So conditioning on $Y \in U$, when $U$ is a linear subspace, is the same as projecting on $U$. $\endgroup$ – kjetil b halvorsen Sep 27 '12 at 14:40
  • $\begingroup$ Sorry but your claim has no sense. $\endgroup$ – Stéphane Laurent Sep 27 '12 at 14:49
  • 1
    $\begingroup$ That is the intuition, a proof maybe must be different. I am out of time now, but note that the multivariate normal distribution can be specified by specifying the (normal) distribution of all linear combinations of the components of $Y$. When the covariance matrix is the projection $P$, choose $u_1, \dots, u_k$ as an orthonormal basis of $U$. $P$ can be written $P=\sum u_i u_i^T $. Choose as coefficient for the linear combination on of the $u_i$, you will see the variance is one. Choose for the coefficient an length-one vector orthogonal to $U$, you will see the variance is zero. $\endgroup$ – kjetil b halvorsen Sep 27 '12 at 16:03
  • $\begingroup$ So the distribution of $P Y$ coincides with the standard normal in $U$, which is the conditional distribution of $Y$ given $Y \in U$. $\endgroup$ – kjetil b halvorsen Sep 27 '12 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.