Lets say I have a uniform random variable $X$ in $[0, 1]$, and another random variable $Y = X$. Clearly, $Y$ is dependent on $X$.

The marginal pdf's for the random variables are $$f_X(t) = f_Y(t) = \begin{cases} 1 && \text{if $t \in [0, 1]$} \\ 0 && \text{otherwise} \end{cases}$$

I am trying to figure out what the conditional pdf for $x$ conditioned on $y$ would be. The cdf seems pretty obvious. $$F_{X|Y}(x|y) = P(X < x | Y = y) = P(y < x) = \begin{cases} 0 && \text{if $x \le y$} \\ 1 && \text{if $x > y$} \end{cases}$$

This is where I am confused. Because the pdf should be the derivative of this function, but we have a jump discontinuity. Thus we should have $$f_{X|Y}(x|y) = 0 \quad\text{if } x \neq y$$

But this seems to break the property $$\int_{-\infty}^{\infty} f_{X|Y}(x|y) dx = 1$$

Can someone help clarify what I am misunderstanding about conditional pdf and cdfs?



If you don't insist on being rigorous, you could say that the conditional pdf is the Dirac delta function, $$ f_{X|Y}(x|y) = \delta( x - y).$$ Roughly speaking, this is an infinite "spike" at $x = y$: the function is zero when $x \neq y$, and takes the value infinity at $x = y$, and its integral over all $x$ is defined to be one.

This fits quite nicely with your discussion about differentiating the cumulative density function. The derivative of the cdf is zero when $x \neq y$, but it is infinite at $x = y$, and the integral of the derivative of the cdf should equal the jump in the cdf at $x = y$, which is indeed one. Again, this statement is not rigorous, because a function with a jump continuity is not really differentiable, and the delta function isn't really a function. (See here and here for more discussion.)

  • $\begingroup$ Thanks. I was trying to come up with some idea like the dirac delta in my head. I am going to read on about that. $\endgroup$ – zrbecker Apr 21 '17 at 23:47

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