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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 $U$.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $U$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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