Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading a proof and it seems like they are using this fact where $(X,Y,Z)$ are discrete random variables):

$p(x,y,z) = p(z)\cdot p(x,y\vert z)$

Where we should understand $p(x,y,z)$ as the probability mass function for the multiple variable $(X,Y,Z)$, $p(z)$ the probability mass function for the random variable $Z$ and $p(x,y\vert z)$ as the probability mass function for the variable $(X,Y\vert Z)$

But is this true? I have tried to derive it using the chain rule, but had no luck.

share|cite|improve this question
Hint: what is the definition of the conditional probability $p(x,y|z)$? – Robert Israel Feb 17 '13 at 20:34
I have deduced that $\frac{p(x,y,z)}{p(z)} = p((x,y)\vert z)$. But is $p((x,y)\vert z) = p(x,y\vert z)$ ? – guestfromthepast Feb 17 '13 at 22:19
I'm not quite sure what you think the distinction is, but presumably both are $P(X = x\text{ and } Y = y | Z = z)$. – Robert Israel Feb 17 '13 at 22:26
Thanks. I interpreted $p((x,y)\vert z)$ as a way of measure the probability for the outcome $(X=x, Y=y)$ when we know that $Z=z$ and $p(x,y\vert z)$ as the probability for the outcome $X=x$ and $Y=y$ when we know $Z=z$. But now I see that the probability is the same :) – guestfromthepast Feb 17 '13 at 22:53

Yes it's true !

On the first side you have:

$$p(x,y,z) = \frac{d}{dx} \frac{d}{dy} \frac{d}{dz} P[X<x, Y<y, Z<z]$$

And on the other:

$$p(x,y|z) = \frac{d}{dx} \frac{d}{dy} P[X<x, Y<y | Z = z]$$

$$ = \frac{d}{dx} \frac{d}{dy} \left( \frac{\frac{d}{dz} P[X<x, Y<y, Z<z]}{\frac{d}{dz} P[Z<z]} \right)$$

$$ = \frac{ \frac{d}{dx} \frac{d}{dy} \frac{d}{dz} P[X<x, Y<y, Z<z]} {p(z)} $$

$$ = \frac{ p(x,y,z) } {p(z)} $$

share|cite|improve this answer
That would be for densities of continuous random variables, not probability mass functions of discrete random variables. – Robert Israel Feb 17 '13 at 20:33
You're right... But for discrete random variables, this is simply the definition of the conditional probabilities? – GHL Feb 17 '13 at 20:43
How do you define $P[X<x,Y<y|Z=z]$ for some continuous random variable $Z$? – Did Feb 18 '13 at 18:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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