conditional probabilityprobabilityprobability theory
Let $X$ and $Y$ be two random variable with density
$f_{X,Y}(x,y)=\begin{cases} \frac{1}{y}, & \text{for } 0<x<y<1, \\[8pt] 0, & \text{otherwise}.\end{cases}$.
To find the conditional distribution of $X$ given that $Y=y$ first I calculated:
$f_Y(y)=\int_0^y \frac{1}{y}dx = 1 $ for $0<y<1 $.
Then $f_{X|Y}(x|y)=\begin{cases} \frac{\frac{1}{y}}{1}, & \text{for } 0<x<y<1, \\[8pt] 0, & \text {otherwise.} \end{cases}$.
Is that correct? I am very unsure about that.
By independence, the joint density function is $f_{X,Y}(x,y) = f_X(x) f_Y(y)$. Be sure to incorporate the information about the support of $f_X$ and $f_Y$ (where it is zero and where it is nonzero).
Similarly, the joint distribution function is $P(X\le x, Y \le y) = P(X \le x) P(Y \le y)$ by independence. Again, be careful: $P(X\le x)$ is piecewise linear.
The joint density looks like this:
So, the marginal distribution of $Y$ is given by the following integral
$$\int_y^1 2 \ dx=2(1-y)$$
if $0\leq y\leq 1$.
Best Answer
Yes, just like for 2 events $A,B$ you have $$\mathbb{P}[A|B] = \frac{\mathbb{P}[A \cap B]}{\mathbb{P}[B]}$$ so too more generally in terms of random variables, $$ f_{X|Y}(x,y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} $$