I have a question in conditional probability. I'm asked to find the conditional distribution, however, I'm unsure about the answer given and would appreciate someone helping straighten out the theory behind it.
Question is from: "A First Course in Probability 9th ed; Sheldon Ross" p.274, Problem 6.42. It asks: The joint density of X and Y is $f(x,y)=c(x^2-y^2)e^{-x}$, and $0\le x\lt \infty, -x\le y\le x$. I am to find the conditional distribution of $Y$, given $X=x$.
I first solved my $f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}$, which is $f_{Y|X}(y|x)=\frac34\frac{x^2-y^2}{x^3}$.
To get the distribution, I need to solve for $F_{Y|X}(y|x)=P(Y\epsilon(-x,x)|X=x)$.
Here is where I get stuck.
The answer given shows that I need to integrate from -x to y... If you're integrating with respect to dy, are you allowed to put y in your limits of integration? I'll type it out so it's clear:
$F_{Y|X}(y|x)=\frac3{4x^3}\int_{-x}^{y}(x^2-y^2)dy$
The final calculation from here is elementary, but like I mentioned, how does this make sense conceptually. Integrating with respect to y doesn't make sense by putting y in the integral...
Thank you kindly in advance.
edit: 1. original equation was changed from + to -, (typo).
2. A change of index was suggested $F_{Y|X}(y|x)=\frac3{4x^3}\int_{-x}^{y}(x^2-t^2)dt$, but I'm still wondering why the integral is not from (-x,x) instead of now (-x,y).
You seem to be on the right track for the conditional density function. $f_{Y\mid X} = \frac{f_{X,Y}(x,y)}{f_X(x)}$
(I haven't checked you integration for the marginal, but it eyeballs right.)
The cumulative density function is $F_{Y|X}(y|x)=P(Y\in(-x,\color{blue}{y})|X=x) = \mathsf P(Y\leq y\mid X=x)$
The other thing is that statements like $\displaystyle\int_{c}^{y} g(y)\operatorname dy = G(y)-G(c)$ are a slight abuse of notation that is unfortunately fairly commonly used.
With a change of index, the definite integral is identical to: $\displaystyle\int_c^y g(t)\operatorname d t = G(y)-G(c)$
So you have: $$\begin{align} F_{Y|X}(y|x) & = \frac3{4x^3}\int_{-x}^{y}(x^2-y^2)dy \\[2ex] & = \frac3{4x^3}\int_{-x}^{y}(x^2-t^2)dt \end{align}$$
Can you continue?