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Let $X$ have density $f_X(x) = 2x$ for $x\in(0,1)$ and zero otherwise. Let $Y$ be uniform on the interval $[1,2].$ Assume that $X$ and $Y$ are independent.

a) Find the joint PDF of $(X,Y)$. Use this to compute $\mathbb P(Y-X\geq3/2).$

b) Find the PDF of $X+Y$.

$\textbf{My Thoughts:}$ For part (a), I have the following. Since both variables are independent, I compute $$f_{XY}(x,y)=f_X(x)f_Y(y)=\begin{cases}2x&\text{if }x\in(0,1)\,\text{and}\,y\in[1,2]\\0&\text{otherwise. }\end{cases}$$

Then I have \begin{align*}\mathbb P(Y-X\geq3/2)&=1-\mathbb P(Y-X<3/2)\\ &=1-\int_{1/2}^{2}\int^{1/2}_{0}2x\,dx\,dy\end{align*}

For part (b) I compute, for $z\in [2,3]$ \begin{align*} f_{X+Y}(z)&=\int_{-\infty}^{\infty}f_{X}(t)f_{Y}(z-t)\,dt\\ &=\int^{1}_{0}2t\,dt\\ &=1, \end{align*} and I have that $f_{X+Y}=0$ otherwise from the above calculation.


Is any of the above correct, especially the bounds on the double integral in part (a)? I am having trouble visualizing the region over which I have to integrate.


Thank you for your time and for any help.

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  • $\begingroup$ If $t \in (0,1)$ then $z-t \in (1,3).$ $\endgroup$ – Dbchatto67 Mar 15 at 10:08
  • $\begingroup$ But whenever $2 \leq z-t < 3$ then $f_Y(z-t)=0.$ Right? So you have consider some cases. $\endgroup$ – Dbchatto67 Mar 15 at 10:10
  • $\begingroup$ @Dbchatto67 The thing I am confused about is that $t$ is a variable, but so is $z$, right? Sorry for my misunderstanding of these inequalities. $\endgroup$ – G the Stackman Mar 15 at 10:13
  • $\begingroup$ observe that $z-t \in [1,2]$ and at the same time $t \in (0,1).$ Then only the integrand is non-vanishing. So $t \in [z-2,z-1]$ and at the same time $t \in (0,1).$ $\endgroup$ – Dbchatto67 Mar 15 at 10:33
  • $\begingroup$ Now if $t \in (0,1)$ then clearly $z-t > 1$ since $z \in [2,3].$ So the only thing to ensure is that whether or not $t \geq z-2.$ $\endgroup$ – Dbchatto67 Mar 15 at 10:37
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First of all for $(a)$ $$\begin{align} \Bbb P \left (Y-X<\frac 3 2 \right ) & = \int_{1}^{\frac 3 2} \int_{0}^{1} 2x\ \text {dx dy} + \int_{\frac 3 2}^{2} \int_{y-\frac 3 2}^{1} 2x\ \text {dx dy}. \\ & = \frac 1 2 + \frac {11} {24}. \\ & = \frac {23} {24}. \end{align}$$ So $$\Bbb P \left (Y-X \geq \frac 3 2 \right ) = 1 - \frac {23} {24} = \frac {1} {24}.$$ Finally for $(b)$ if $Z=X+Y$ and $f_Z(z)$ denotes the PDF of $Z$ then $$ f_Z(z) = \left\{ \begin{array}{ll} z^2 - 2z + 1 & \quad 1 < z < 2 \\ 4z - z^2 - 3 & \quad 2 \leq z < 3 \\ 0 & \quad \text {elsewhere} \end{array} \right. $$

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For (a) see the answer of DBchatto67.

Let $\left[x+y\leq z\right]$ denote the function $\mathbb{R}^{3}\to\mathbb{R}$ that takes value $1$ if $x+y\leq z$ and takes value $0$ otherwise.

Then for the CDF of $Z:=X+Y$ we find that

$\begin{aligned}F_{Z}\left(z\right)=P\left(X+Y\leq z\right) & =\int\int\left[x+y\leq z\right]f_{X}\left(x\right)f_{Y}\left(y\right)dydx\\ & =\int_{0}^{1}\int_{1}^{2}\left[x+y\leq z\right]2xdydx\\ & =2\int_{0}^{1}x\int_{1}^{2}\left[y\leq z-x\right]dydx \end{aligned} $

Now we discern cases.

  • If $z\leq1$ then $P\left(X+Y\leq z\right)=0$.

  • If $1<z<2$ then

$\begin{aligned}P\left(X+Y\leq z\right) & =2\int_{0}^{z-1}x\int_{1}^{z-x}dydx\\ & =2\int_{0}^{z-1}x\left(z-x-1\right)dx\\ & =\frac{1}{3}\left(z-1\right)^{3} \end{aligned} $

  • If $2\leq z<3$ then

$\begin{aligned}P\left(X+Y\leq z\right) & =2\int_{0}^{1}x\int_{1}^{\min\left(2,z-x\right)}dydx\\ & =2\int_{0}^{z-2}x\int_{1}^{2}dydx+2\int_{z-2}^{1}x\int_{1}^{z-x}dydx\\ & =2\int_{0}^{z-2}xdx+2\int_{z-2}^{1}x\left(z-x-1\right)dx\\ & =z-\frac{5}{3}-\frac{1}{3}\left(z-2\right)^{3} \end{aligned} $

  • If $z\geq3$ then $P\left(X+Y\leq z\right)=1$

We find the PDF by differentiating the CDF

$f_{Z}\left(z\right)=\left(z-1\right)^{2}$ if $1<z<2$ and $f_{Z}\left(z\right)=1-\left(z-2\right)^{2}=\left(3-z\right)\left(z-1\right)$ if $2\leq z<3$ and $f_{Z}\left(z\right)=0$ otherwise.

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