I have two random variables, $X, Y$ and their joint pdf, $f_{XY}(x,y )$.

I am able to find the marginal PDFs, $f_X(x)$ and $f_Y(y)$ using $f_X(x) = \int_{-\infty}^{\infty}f_{XY}(x,y)dy$ and similar for $f_Y(y)$.

Now I am given a new random variable: $Z = g(X, Y) = X - Y$

I found $f_Z(z)$ by first finding $F_Z(z)$ and taking its derivative with respect to $z$

$$ \begin{align} F_Z(z) = \mathbb{P}(Z \le z) \\ = \mathbb{P}(X - Y \le z) \\ = \mathbb{P}(Y \ge X - z) \\ = 1 - \mathbb{P}(Y \le X - z) \\ = 1 - \iint{f_{XY}(x,y)dydx} \end{align} $$ (The limits of integration in double integral are functions of $z$ (or constants).)

Then $f_Z(z) = \frac{d}{dz}F_Z(z)$

Now I need to find some conditional probabilities, such as $\mathbb{P}(X - Y > a | Y < a)$ and $\mathbb{P}(X - Y = b | Y < a)$.

Here is where I am stuck. If they were "simple" conditionals without a function of $X$ and $Y$, e.g., $\mathbb{P}(X < b|Y < a)$, I could just find the conditional pdf $f_{X|Y}(x|y) = \frac{f_{XY}(x,y)}{f_Y(y)}$ since I know the joint pdf of $X$ and $Y$ and the marginal pdfs.

But how would I proceed to find $\mathbb{P}(X - Y > a | Y < a)$? Is there some way to find a joint PDF of $Z$ (which is $X - Y$) and $Y$?

  • $\begingroup$ I should add: $X$ and $Y$ are not independent. $f_{XY}(x,y) \ne f_X(x)f_Y(y)$ $\endgroup$ – tony_tiger Mar 17 '14 at 6:39
  • $\begingroup$ Might I need to make some kind of substitution? $X = Y + Z$ Then could I write $f_{ZY} = f_{XY}(y+z, y)$ ? $\endgroup$ – tony_tiger Mar 18 '14 at 18:57

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