Marginal Densities I just have a few questions about joint density and marginal density questions.
Q1: Joint Distribution $f_1=2x+4y$ on triangle with vertices $(0,0), (0,1),(1,0)$.
Sketch the region and compute marginal density of $X$. Use this to compute the expected value of $X$. Find the conditional density of Y given X. If X is .5, what is the expected value of $Y$. Lastly, the total is $Z=X+Y$. Find the density of Z.
The region is clearly just the triangle with those vertices. The marginal density is the integral of $f_1$ from $(0)$ to $(1-x)$ $dy$. The expected value is simply integral from 0 to 1 of $xf_{1x}(X)=\frac{1}{3}$.
The conditional density is simply the joint density divided by the marginal density of $X$. This I found to be  $$\frac{2x+4y}{2-2x}=\frac{x+2y}{1-x}$$
Therefore $E(Y|X=.5)=\int^{.5}_0 yf_{y|x}=\frac{7}{24}$
Lastly, the cumulative distribution function of $X+Y$ I determined as $$\int_0^z\int^{z-x}_0f(x,y)dydx=\int_0^z\int^{z-x}_0(2x+4y)dydx=\int_0^z((2xy+2y^2))|^{z-x}_0=\int_0^z2x(z-x)+2(z-x)^2$$
$$=\int_0^z2xz-2x^2+z^2-2zx+x^2=\int_0^z-x^2+z^2=(\frac{-x}{3}+xz^2)|^z_0=\frac{-z^3}{3}+z^3$$.
Therefore the density is simply the derivative of this, which I found to be $-z^2+3z^2$
Q2: $X$ is the percentage spent on total budget. $Y$ is spend on books. Joint Distribution is $f_2=8xy$ for $0\le x\le 1$ and $0\le y \le x$. Find the marginal densities and sketch both. then find their expected values. Find the conditional density of $f{Y|X}$ and evaluate the $E(Y|X=.6)$. Let $Z=Y/X$. Find the PDF or CDF of $Z$ and find the expected value of $Z$.
Marginal Density of $X$:$\int^x_0 8xydy=4x^3$
Marginal Density of $Y$:$\int^1_y 8xydx=4y-4y^3$
$E(X)=\int^1_0xf_xdx=\int^1_04x^4dx=\frac{4}{5}$
$E(Y)=\int^1_0yf_ydy=\int^1_04y^2-4y^4=\frac{8}{15}$  
The conditional density of $f{Y|X}=\frac{8xy}{4x^3}=\frac{4y}{x^2}$
$E(Y|X=.6)=\int^.6_0 y(\frac{4y}{.6^2})=\frac{4}{5}$
Lastly, $$\int^1_0\int^{zx}_08xydydx=\int^1_04xy^2|^{zx}_0=\int^1_04z^2x^3=z^2x^4|^1_0=z^2$$
Thus by taking the derivative the density of $z$ is $2z$.
$E(Z)=\int^1_0 z*2z=\frac{2}{3}$

It looks like you have already worked out parts of it but I'll include all the working anyway.
Q1.
(a) Find $f_X(x)$:
\begin{eqnarray*}
f_X(x) &=& \int_0^{1-x} (2x+4y)\;dy \\
&=& \left[ 2xy+2y^2\right]_0^{1-x} \\
&=& 2x(1-x)+2(1-x)^2 \\
&=& 2-2x.
\end{eqnarray*}
(b) Find $E(X)$:
\begin{eqnarray*}
E(x) &=& \int_0^{1} xf_X(x)\;dy \\
&=& \int_0^{1} (2x-2x^2)\;dy \\
&=& \left[ x^2-\dfrac{2}{3}x^3 \right]_0^{1} \\
&=& 1-\dfrac{2}{3} \\
&=& \dfrac{1}{3}.
\end{eqnarray*}
(c) Find $f_{Y|X}(y|x)$:
\begin{eqnarray*}
f_{Y|X}(y|x) &=& \dfrac{f_{X,Y}(x,y)}{f_X(x)} \\
&=& \dfrac{2x+4y}{2-2x} \\
&=& \dfrac{x+2y}{1-x} \qquad\qquad\text{for $0\leq x,\quad 0\leq y,\quad x+y\leq 1$.}
\end{eqnarray*}
(d) Find $E(Y\mid X=0.5)$:
\begin{eqnarray*}
E(Y\mid X=0.5) &=& \int_0^{0.5} yf_{Y|X}(y|0.5)\;dy \qquad\qquad\text{(upper limit $=1-x=1-0.5=0.5$)} \\
&=& \int_0^{0.5} \dfrac{y/2+2y^2}{1-1/2}\;dy \\
&=& \int_0^{0.5} (y+4y^2)\;dy \\
&=& \left[ \dfrac{1}{2}y^2+\dfrac{4}{3}y^3 \right]_0^{0.5} \\
&=& \dfrac{1}{8} + \dfrac{1}{6} \\
&=& \dfrac{7}{24}.
\end{eqnarray*}
(e) Find $f_Z(z)$ where $Z=X+Y$. Note that this involves a single integral, not a double integral.
\begin{eqnarray*}
f_Z(z) &=& \int_0^{z} f_{X,Y}(x,z-x)\;dx \\
&& \qquad\text{(upper limit $z$ is the largest possible $X$ given that $X+Y=z$)} \\
&=& \int_0^{z} (2x+4(z-x))\;dx \\
&=& \int_0^{z} (4z-2x)\;dx \\
&=& \left[ 4zx-x^2 \right]_0^{z} \\
&=& 3z^2.
\end{eqnarray*}
Q2. Finding $f_X, f_Y, E(X), E(Y)$ can be done the same way as in Q1.
(a) Find $f_Z$. I'll use the "change of variable" method.
We have $Z=Y/X$. Also let $W=X$. Then $Y=WZ$ and $X=W$. Therefore our Jacobian is:
$$J=\begin{vmatrix}
        \dfrac{\partial x}{\partial z} & \dfrac{\partial y}{\partial z} \\
        \dfrac{\partial x}{\partial w} & \dfrac{\partial y}{\partial w} \\
        \end{vmatrix}
=\begin{vmatrix}
        0 & w \\
        1 & z \\
        \end{vmatrix} = -w.
$$
Therefore,
\begin{eqnarray*}
f_{Z,W}(z,w) &=& f_{X,Y}(x(z,w),\;y(z,w))\;|J| \\
&=& 8zw^3 \qquad\text{for $0\lt z \lt 1\;$ and $\;0\lt w \lt 1$}. \\
\end{eqnarray*}
Now,
\begin{eqnarray*}
f_{Z}(z) &=& \int_{w=0}^{1} f_{Z,W}(z,w)\;dw \\
&=& \int_{w=0}^{1} 8zw^3 \;dw \\
&=& \left[ 2zw^4 \right]_{w=0}^{1} \\
&=& 2z.
\end{eqnarray*}
(b) Find $E(Z)$:
\begin{eqnarray*}
E(Z) &=& \int_{0}^{1} zf_Z(z)\;dz \\
&=& \int_{0}^{1} 2z^2 \;dz \\
&=& \left[ \dfrac{2}{3}z^3 \right]_{0}^{1} \\
&=& 2/3.
\end{eqnarray*}
