# Finding joint density function of two independent random variables

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.

• If $t \in (0,1)$ then $z-t \in (1,3).$ Commented Mar 15, 2019 at 10:08
• But whenever $2 \leq z-t < 3$ then $f_Y(z-t)=0.$ Right? So you have consider some cases. Commented Mar 15, 2019 at 10:10
• @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. Commented Mar 15, 2019 at 10:13
• 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 $t \in (0,1).$ Commented Mar 15, 2019 at 10:33
• 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 $t \geq z-2.$ Commented Mar 15, 2019 at 10:37

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 $$f_Z(z) = \left\{ \begin{array}{ll} z^2 - 2z + 1 & \quad 1 < z < 2 \\ 4z - z^2 - 3 & \quad 2 \leq z < 3 \\ 0 & \quad \text {elsewhere} \end{array} \right.$$

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 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 $$1 and $$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.