find $P(Y|X)$ from $P(X=x,Y\leq y)$

Question

Suppose $$P(X=n,Y\leq y) = b\int_0^y \frac{(at)^n}{n!}e^{-(a+b)t}\,dt.$$ How do you actually find $P(Y|X)$? I tried from bayes that $P(Y=y|X=x)=\frac{d}{dy}\frac{P(X,Y\leq y)}{P(X)}$ and after finding the integral, it seems that differentiating w.r.t y does not simplify it at all.

Answer 1

By definition the conditional CDF is: $\mathsf P(Y\leq y\mid X=n)=\dfrac{\mathsf P(X=n, Y\leq y)}{\mathsf P(X=n)}$

Also recall that by Law of Total Probability: $\mathsf P(X=n)=\lim\limits_{y\to\infty}\mathsf P(X=n, Y\leq y)$

So indeed

$$\begin{align}\dfrac{\mathrm d ~~}{\mathrm dy}\mathsf P(Y\leq y\mid X=n)&=\dfrac{\mathrm d ~~}{\mathrm d y}\dfrac{\int\limits_0^y \frac{b}{n!}(at)^n\mathrm e^{-(a+b)t}\mathrm d t}{\int\limits_0^\infty \frac{b}{n!}(at)^n\mathrm e^{-(a+b)t}\mathrm d t}\\[2ex]&=\dfrac{\dfrac{\mathrm d ~~}{\mathrm d y}\int\limits_0^y t^n\mathrm e^{-(a+b)t}\mathrm d t}{\int\limits_0^\infty t^n\mathrm e^{-(a+b)t}\mathrm d t}\end{align}$$

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Now all you need is that by the Fundamental Law of Calculus: $f(y)=\dfrac{\mathrm d }{\mathrm d y}\int\limits_0^y f(t)\mathrm d t$ , and that the factorial of a natural number $n$ is: $n!=\int\limits_0^\infty z^n\mathrm e^{-z}\mathrm d z$.

Answer 2

Let $N=X$ so $P(Y|N)=\frac{\int\limits_0^y\frac{(at)^N}{N!}e^{-(a+b)t}dt}{\int\limits_0^\infty\frac{(at)^N}{N!}e^{-(a+b)t}dt}$