The joint density function of two continuos random variables $X$ and $Y$ is given by:
$f(x,y) = 8xy$ if $0\le y\le x\le 1$ and $0$ otherwise.
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Calculate $P(X \le \frac{1}{2})$
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Calculate $P(Y \le \frac{1}{4} \mid X = \frac{1}{2})$
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Calculate the expected value of $Y^3$ if $X = \frac{1}{2}$
I would just like to check whether I am solving these questions in the right way. For question a), I think you first need to derive the marginal density function for $X$. However, I am unsure whether I obtain this by integrating over from $0$ to $x$ or from $0$ to $1$ (which one is correct and why?). Also, I wasnt entirely sure about how to do b, could anyone show me how that probability would be obtained?.
I think I can do c, however, for it to be correct, I first need te correct answer to question a. Could anyone please help me out?
Best Answer
(1) The marginal PDF of $X$ is $$f_X(x)=\int_{-\infty}^\infty f(x,y)\,dy\\ =\begin{cases}\int_0^x 8xy\,dy & ,\text{if} \,\,\, 0<x<1\\0 & \text{otherwise}\end{cases}$$ The marginal PDF of $Y$ is $$f_Y(y)=\int_{-\infty}^\infty f(x,y)\,dx\\ =\begin{cases}\int_y^1 8xy\,dx & ,\text{if} \,\,\, 0<y<1\\0 & \text{otherwise}\end{cases}$$
(2) The conditional PDF of $Y|X=x$ is $$f_{Y|X}(y|x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}\\ =\begin{cases}\dfrac{8xy}{\int_0^x 8xy\,dy} & ,\text{if} \,\,\, 0<y<x\\0 & \text{otherwise}\end{cases}\\=\begin{cases}\dfrac{2y}{x^2} & ,\text{if} \,\,\, 0<y<x\\0 & \text{otherwise}\end{cases}$$ So $P(Y \le \frac{1}{4} \mid X = \frac{1}{2})=\int_0^{1/4}\dfrac{2y}{(1/2)^2}\,dy$