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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.

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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

1 Answer 1

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:

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

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

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

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

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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

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