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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 $Y$.


(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

$\int_0^1 \int_x^{1-x}2e^{-(x+y)}dydx$




$e^0 + 0 - e^{-2} + 2e^{-1} = 1-\frac{1}{e^2}+\frac{2}{e}$

Hence $Pr(x+y<1) = 1-\frac{1}{e^2}+\frac{2}{e}$

(f) Unsure how to do.

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up vote 1 down vote accepted

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.

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When you integrated the $2e^{-x}e^{-y}$ where did the minus go to give $2e^{-2x}$? I tried using $y=x$ rather than $y=0$ as a limit, but as I had the integration with -2 not 2 i was getting $2=-2$ and was confused yet didn't want to assume the integration should have been $2$ not $-2$ – Matt Oct 20 '13 at 19:02
Apart from the term $2e^{-x}$, the antiderivative is $-e^{-y}$. Plug in "$\infty$", take away the result of plugging in $x$. That cancels the minus sign. Anyway, your minus cannot be right, densities are non-negative. – André Nicolas Oct 20 '13 at 19:06
For f), using my notes for $Cov(x,y)$ i got $1/4$. To do this I used your formula and got to $\int_0^yx[2e^{-x}]dx$ and said $x[2e^{-x}]$ is the same as $2E[x]$ because in my notes it said $\int_{x_1=0}^{x_1=\infty}x_1(\lambda x_1e^{-\lambda x_1}+e^{-\lambda x_1})dx_1$ is the same as $E[x_1^2] + \frac{E[x_1]}{\lambda}$. Is this the same and then correct? – Matt Oct 22 '13 at 18:57
My integration is unreliable, I have only done it once. However, I get $E(XY)=1$, $E(X)=\frac{1}{2}$ (obvious, exponential parameter $2$) and $E(Y)=1.5$. That gives covariance $1-(1/2)(1.5)=0.25$, same as yours. – André Nicolas Oct 22 '13 at 19:18

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