[Math] Finding the conditional pdf of $Y$ given $X=x$ from a joint pdf. Answer confirmation!

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I have a continuous joint pdf, and I am working out the conditional pdf of Y given X=x.

Is my method correct?

I am given: $f_{X,Y}(x,y) = x\cos y , 0 \lt x \lt \frac{\pi}{2}, 0 \lt y \lt x$

$$\text{I found marginal pdf of X: } f_X(X=x) = \int_0^x x \cos y \, \mathrm{dy} = [x \sin y]_0^x = x \sin x$$

$$\text{Then using the following : }Pr(Y=y \mid X = x) = \frac{f_{X,Y}(x,y)}{f_X(x)}= \frac{x \cos y}{x \sin x}$$ $$\text{giving the conditional pdf of } Y \text{ given } X = x, \rightarrow\frac{\cos y}{\sin x}$$

Is this correct? Thank you for your time!

Assuming above is correct, how would I then go about calculating $Pr(2X + Y \geq \frac{\pi}{2})$ or $Pr(X + Y \geq \frac{\pi}{2})$?

It sounds like I want for $Pr(X + Y \geq \frac{\pi}{2})$ to integrate $X+Y$ from $\frac{\pi}{2}$ to $\infty$ with respect to $x$ perhaps, but where does my $f_{Y \mid X}(y \mid x)$ come in?

Best Answer

Your computations are correct you just need to specify the domain of the conditional pdf of $Y$ given $X$ and say whereit is zero.

For $Pr(2X+Y\geq \pi/2)$, you need to compute the distribution functions of the random variables $U=2X$ and then $V=U+Y$ using convolution product (assuming that $U$ and $Y$ are independent).