why is this conditional PDF of a random variable a delta function... or not?

131 Views Asked by At

I was reading page 11 of an introduction to random matrices; in particular, this piece:

description of problem

Why is (2.2) true? It seems reasonable, but I think this delta function is confusing me. I want to break the equation down into probabilities of random variables "equaling" values. For example,

$$ P(X_1 \in [x_1, x_1 + dx_1], X_2 \in [x_2, x_2 + dx_2], X_3 \in [x_3, x_3 + dx_3]) = dx_1 dx_2 dx_3 \frac{e^{-\frac{1}{2}x_1^2}}{\sqrt{2\pi}} \frac{e^{-\frac{1}{2}x_2^2}}{\sqrt{2\pi}} \frac{e^{-\frac{1}{2}x_3^2}}{\sqrt{2\pi}} $$

(though I might just write this as $P(X_1 = x_1, X_2 = x_2, X_3 = x_3)$ for shorthand). Note that the above does not seem exactly consistent with the authors' notation since they seem to use the same symbol for the random variable and the value, but anyways.

My thought right now is that "multiplying" (2.2) by $ds$ we basically have the law of total probability, where $p(s)ds = P(S = s)$ and then

$$ P(S = s | X_1 = x_1, X_2 = x_2, X_3 = x_3) = \delta\left(s - \sqrt{(x_1 - x_2)^2 + 4x_3^2}\right) \tag{A}\label{A}$$

so that (2.2) more or less amounts to writing

$$ P(S = s) = \sum_{x_1, x_2, x_3} P(X_1 = x_1, X_2 = x_2, X_3 = x_3) P(S = s | X_1 = x_1, X_2 = x_2, X_3 = x_3) . $$

Is this all how you would think about why (2.2) is true "by definition"? And if so, why is it that equation \eqref{A} is true?

1

There are 1 best solutions below

0
On

Since $s=\sqrt{(x_1-x_2)^2+4x_3^2}$, the $\delta$ factor you've encountered is its conditional probability density. (I don't call it a probability density function, because technically it's not a function.) To recover the density of $s$ when the $x_i$ are random, we do what you said: multiply by the density in $x$-space, then integrate over $x$-space. If we use lower-case $p$s to denote densities rather than probability masses, the conditional densities satisfy $p(s)=\int d^3xp(x)p(s|x)$, in analogy with a discrete problem where upper-case $P$ masses satisfy $P(s)=\sum_xP(x)P(s|x)$.