Can someone please check if the below answers are correct? Preparing for an exam and since I was so naive to sign up for upper level courses that are using Calculus (which I never studied), I'm struggling a bit with making sure that all of my answers are correct and that I am, in fact, able to solve the problems given. Thank you in advance!
P.S. I'm self studying, hence, why I can't check if my answers are right.
Let $X,Y$ be continous random variables, where $X$ and $Y$ take values in the intervals $[0,1/2]$ and $[0,1]$, respectively. The joint density is given by: $f_{X,Y}(x,y) = 16xy$ for $0\leq x\leq 0.5$ and $0\leq y\leq 1$
a) Determine the $P(X\leq 0.25,Y\geq 0.5)$.
\begin{align*}
P(X\leq 0.25,Y\geq 0.5) & = \int_{0.5}^{1}\int_{0}^{0.25}16xy\mathrm{d}x\mathrm{d}y = \int_{0.5}^{1}[8x^{2}y]_{x=0}^{x=0.25}\mathrm{d}y\\
& =\int_{0.5}^{1}[0.5y]\mathrm{d}y = [0.25y^{2}]_{y=0.5}^{y=1}\\
& = 0.25-0.0625 = 0.1875
\end{align*}
b) Determine the marginal density of $X$.
Marginal PDF of $X$ is:
\begin{align*}
f_{X}(x) = \int_{0}^{0.5} 16xy\mathrm{d}y = [8xy^{2}]_{y=0}^{y=0.5} = 2x
\end{align*}
c) Calculate the expected value of $X$.
\begin{align*}
\textbf{E}(X) & = \int_{0}^{0.5}xf(x)\mathrm{d}x = \int_{0}^{0.5}x2x\mathrm{d}x = \int_{0}^{0.5}2x^{2}\mathrm{d}x\\
& = [2/3x^{3}]_{x=0}^{x=0.5} = 0.08333
\end{align*}
Best Answer
As far as I have checked, the answer to (a) is correct. However, the answer given to (b) is wrong, since the upper integration limit is wrong: it should be one instead of one half, which results into the marginal probability density function $f_{X}(x) = 8x$. Once the second answer is wrong, the same applies to the last answer, since it is based on the second. But I think you can proceed from my tips.