I am unsure with question c/d/e/f but I will give my answers for all questions I have attempted.
The random variables $X$ and $Y$ have joint probability density function
$f_{X,Y}(x,y) = ke^{-(x+y)},\ x>0,\ y>0,\ 0<x<y.$
(a) Sketch the region over which the joint probability density function is non-zero.
(b) Show that $k = 2.$
(c) Find the marginal Distribution of $X$; name the distribution and state its mean and variance.
(d) Find the conditional distribution of $Y|X = x$ for $x>0$. Pay particular attention to the range of $Y$ conditional on $x$.
(e) Calculate $Pr(X + Y<1)$.
(f) Given that $E(Y) = 1.5$ and $Var(Y) = 1.25$, calculate the correlation between $X$ and $Y$.
ANSWERS:
(a) Don't know how to draw on here but it's the area above $y=x$ and $y>0$.
(b) Here I used double integration
$\int_0^\infty \int_0^y ke^{-(x+y)}dxdy = 1$
$\int_0^\infty [-ke^{-(x+y)}]_0^y dy = \int_0^\infty[-ke^{-y}+ke^{-2y}]dy$
$[ke^{-y}-\frac{k}{2}e^{-2y}]_0^\infty = [ke^{-0} – \frac{k}{2}e^{-2\cdot0}]-[ke^{-\infty}-\frac{k}{2}e^{-2\cdot \infty}] = [k – \frac{k}{2}]-[0] = \frac{k}2$
Hence $\frac{k}{2}=1$ and $k=2$
(c) $P_x(x)=\int_yP_{X,Y}(x,y)dy=\int_0^\infty2e^{-(x+y)}dy$
$[-2e^{-(x+y)}]_0^\infty= -2e^{-x}$
Presuming that the integration I have done is correct the marginal distribution of $X$ is $-2e^{-x}$ and hence this is a gamma distribution as the limits are $(0,\infty)$ and it just doesn't match the binomial distribution either.
Gamma distribution: $\frac{1}{\Gamma(\alpha)\cdot \beta^\alpha}\cdot x^{\alpha-1}\cdot e^{-\frac{x}{\beta}}$
Hence $\frac{1}{\Gamma(\alpha)\cdot \beta^\alpha}\cdot x^{\alpha-1}\cdot e^{-\frac{x}{\beta}} = -2e^{-x}$
This is where I have gone wrong as using $\beta = 1$ to leave $e^{-x}$ as that i then need to find:
$\frac{1}{\Gamma(\alpha)\cdot 1^\alpha}\cdot x^{\alpha-1} = -2$ and using $\alpha =1$ to make $x^0$ and getting rid of $x$ but this makes $1=-2$.
(d) unsure how to do this question also.
(e) Adding the line $y=1-x$ to the graph from *(a)*we get a new region to find
$\int_0^1 \int_x^{1-x}2e^{-(x+y)}dydx$
$\int_0^1[-2e^{-(x+y)}]_x^{1-x}dx$
$\int_0^1[(-2e^{-2x})-(-2e^{-1})]dx$
$[e^{-2x}+2xe^{-1}]_0^1$
$e^0 + 0 – e^{-2} + 2e^{-1} = 1-\frac{1}{e^2}+\frac{2}{e}$
Hence $Pr(x+y<1) = 1-\frac{1}{e^2}+\frac{2}{e}$
(f) Unsure how to do.
Best Answer
On (c), we want to "integrate out" $y$. So the density function of $X$ is $$\int_{y=x}^\infty 2e^{-x}e^{-y}\,dy.$$ Integration yields $2e^{-x}e^{-x}$. So $X$ has exponential distribution, parameter $2$.
On (d), recall that the conditional density of $Y$ given that $X=x$ is $$\frac{f_{X,Y}(x,y)}{f_X(x)}.$$ Note that the conditional density is $0$ for $y\lt x$.
On (e), the setup is good. There is a small sign error in the calculation. The number obtained is impossible, it is greater than $1$.
On (f), we will first need the covariance, that is, $E(XY)-E(X)E(Y)$. For that, we need to find $$\iint_D (xy) 2e^{-(x+y)}\,dydx,$$ where $D$ is the region where our joint density is positive.