A correct transformation that's wrong

Question

I bumped into the following problem and can't see how I am supposed to calculate divide the intergral into correct areas of definition.

The problem is as follows: $X$ and $Y$ are random variables with joint density as

$$f_{X,Y} = \begin{cases}(2/5)(2x+3y) & 0 < x, y < 1 \\ 0 & \text{o.w.} \end{cases},$$

I want to calculate the distribution of $2X+3Y$.

So I did the following:

Set $U = X$ and $V = 2X + 3Y$ from the latter I derived $Y = (V - 2U)/3$. Here I see that if $X$ and $Y$ are defined in $[0,1]$ then $V$ is defined in $[0,5]$.

Now: $$\begin{align*} f_v(v) &= \int f_{U,V}(u,v) du \\ &= \int f_{X,Y}\left(u, \frac{v - 2u}{3}\right) |J| du \\ &= \frac{2}{15} \int 2u + 3\cdot\frac{v - 2u}{3} du \\ &=\frac{2 v}{15} \int du \\ &= \frac{2 v}{15} \left[ \, u \, \big|_0^v \, \right] \\ &= \frac{2v^2}{ 15} \end{align*}$$

But the answer is something else.

Answer 1

The method you want to use for the sum of random variables is called convolution. It goes like this:

You want

$$f_{2X + 3Y}(z) = P(2X + 3Y = z) = P\left(X = x, Y = \dfrac{z - 2x}{3}\right) = f_{X,Y}\left(x, \dfrac{z - 2x}{3}\right)$$

for a given $z$. You are right that $0 < 2X + 3Y < 5$. So, using the joint distribution of $X$ and $Y$ to get

$$ \begin{align*} f_{2X + 3Y}(z) &= \int_{0}^{5} f_{X,Y}\left(x, \dfrac{z - 2x}{3}\right) \, dx \\ \\ &= \frac{2}{5} \int_{0}^{5} 2x + 3\left(\frac{z-2x}{3}\right) \, dx \\ \\ &= \frac{2}{5} \int_{0}^{5} 3z \, dx \\ \\ &= 6z. \end{align*}$$

Answer 2

You were on the right track until failing to properly transform the conditions $0<x<1,\ 0<y<1$.

We have $$f_{X,Y}(x,y)={2\over 5}(2x+3y)[0<x<1][0<y<1]$$ where $[...]$ are Iverson brackets.

Therefore, with $U=X$ and $V=2X+3Y$, we have $$\begin{align}&f_{U,V}(u,v)\\[2mm] &=f_{X,Y}\left(u, {v - 2u\over 3}\right) |J|\\[2mm] &={2 \over 5}v\,[0<u<1]\left[0<{v - 2u\over 3}<1\right]\,{1\over 3}\\[2mm] &={2 \over 15}v\,[0<u<1]\left[{v - 3\over 2}<u<{v\over 2}\right]\\[2mm] &={2 \over 15}v\left\{ \underbrace{[0<v\le 2]\left[0<u<{v\over 2}\right]}_{A}+ \underbrace{[2<v\le 3][0<u<1]}_{B}\\ + \underbrace{[3<v\le 5]\left[{v-3\over 2}<u<1\right]}_{C}\right\}\\[2mm] \end{align}$$ where we've written the conditions on $(u,v)$ in a form convenient for integration over $u$ to obtain the marginal density for $V$. The terms labelled $A,B,C$ describe disjoint regions in the $(u,v)$-plane. Here's a sketch to make the limits of integration more obvious (with apologies for the poor image quality):

$A\cup B\cup C$ is a parallelogram, with $B$ a unit square sandwiched between triangles $A$ and $C$.

Thus, $$\begin{align}f_V(v) &= \int f_{U,V}(u,v) du \\ &={2 \over 15}v\left\{ [0<v\le 2]\int\left[0<u<{v\over 2}\right]\,du+ [2<v\le 3]\int[0<u<1]\,du\\ + [3<v\le 5]\int\left[{v-3\over 2}<u<1\right]\,du\right\}\\[2mm] &={2 \over 15}v\left\{ [0<v\le 2]{v\over 2}+ [2<v\le 3]1+ [3<v\le 5]{5-v\over 2}\right\}\\[2mm] \color{blue}{f_V(v)}&\color{blue}{={1 \over 15}v^2[0<v\le 2]+ {2\over 15}v[2<v\le 3]+ {v(5-v)\over 15}[3<v\le 5]}.\\[2mm] \end{align}$$

As a check, we can see that this integrates properly to $1$: $$\begin{align}\int f_V(v)\,dv&={1 \over 15}{1 \over 3}[v^3]_0^2 + {2\over 15}{1 \over 2}[v^2]_2^3+ {1\over 15}[(5 v^2)/2 - v^3/3]_3^5\\[2mm] &={8\over 45}+{1\over 3}+{22\over 45}\\[2mm] &=1 \end{align}$$