1
$\begingroup$

I have random variables X and Y. X is chosen randomly from the interval (0,1) and Y is chosen randomly from (0, x). I want to calculate the conditional PDF of Y given X. I want to do this by calculating the joint PDF of X and Y and dividing that by the marginal PDF of X.

The marginal PDF of X is simply 1, since we're equally likely to pick a number from the range of (0,1). We can verify this using calculus by taking the derivative of the CDF, which is simply F(X <= x) = x/1, or x. The derivative of xdx = 1.

I'm struggling with the joint PDF. I have a strong feeling it's 1/x, however I'd like some advice on how to get it.

$\endgroup$
2
$\begingroup$

Yes.   The density function for a random variable uniformly distributed over support $(0;1)$ is: $$f_X(x)= \mathbf 1_{x\in(0;1)}$$

Then, assuming that $Y$ is uniformly selected in the interval $(0;X)$ , the conditional probability density function is, indeed, $$f_{Y\mid X}(y\mid x) = \frac 1 x \; \mathbf 1_{y\in(0;x), x\in(0;1)}$$

The joint density function is: $f_{X,Y}(x,y) = f_X(x) \; f_{Y\mid X}(y\mid x)$

And the marginal of Y, is $f_Y(y)=\int_0^1 f_{Y\mid X}(y\mid x)\;f_X(x)\,\operatorname d x$

You can do the rest.

$\endgroup$
  • $\begingroup$ thanks, this was very helpful $\endgroup$ – A user Jun 25 '15 at 4:00
1
$\begingroup$

The conditional pdf of $Y$ given $X$ is given to you. Choose $X$ from $(0,1)$, then choose $Y$ from $(0,X)$. It is telling you that given $X$, you know the distribution of $Y$.

$$ f_{Y\mid X}(y \mid x) = 1_{[0,x]} \cdot \frac{1}{x} $$

for all $x,y \in (0,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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