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.
Best Answer
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.