Understanding solution to Bain, Engelhardt 6.14; Finding distribution of sum of random variables

Question

I am trying to understand a solution to the following problem, which is 6.14 from Bain and Engelhardt's Introduction to probability and mathematical statistics. My question comes after the solution.

Problem statement: Let $X$ and $Y$ have joint pdf $f(x, y) = 4e^{-2(x+y)}; 0<x<\infty, 0<y<\infty,$ and zero otherwise.

(a) Find the CDF of $W = X + Y$.

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Solution: Let $V=X$. Then we have the system $$\begin{cases} x=v \\y=w-v\end{cases}$$ The corresponding Jacobian transformation is $$J = \begin{pmatrix} \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$$ with determinant $$\det J = 1.$$ So the joint density of $W$ and $V$ is \begin{align}f_{W, V}(w, v) & = f_{X, Y}(x=v; y=w-v) \times \det J \\ & = 4e^{-2(v+w-v)} \times 1 \\ & = 4e^{-2w}. \end{align} Now we check for the limits \begin{align} v>0, w-v>0 & \implies \\ v>0, w>v & \implies \\ f_{W,V}(w,v) = 4e^{-2w}; 0<v<w. \end{align} Next we will get the marginal density of $W$ by integrating \begin{align} f_W(w) & = \int_{0}^{w} 4e^{-2w}dv \\ & = 4e^{-2w}[v]_{0}^{w} \\ & = 4we^{-2w}. \end{align} So we have $$f_W(w) = \begin{cases} 4we^{-2w} & ;w>0 \\ 0 & ; \text{otherwise} \end{cases}. $$ Therefore the distribution of W is given by \begin{align} F_W(w) & = \int_{0}^{w} 4we^{-2w} dw \\ & = 1-e^{-2w}-2we^{-2w}. \end{align}

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My question: I don't understand the very first step, where they define $V=X$. How is this allowed? I understand that the purpose of this is to create a one-to-one transformation between $(X, Y)$ and $(W, V)$, but...it just seems to come out of nowhere. What makes this step valid?

Answer 1

As a general principle, when $\langle X, Y\rangle$ has a bijection with $\langle W, V\rangle$ such that $X=g_{\small x}(W,V)$ and $Y=g_{\small y}(W,V)$, then: $$f_{\small W,V}(w,v)=f_{X,Y}(g_{\small x}(w,v),g_{\small y}(w,v))\cdot\left\lVert\dfrac{\partial\langle g_{\small x}(w,v),g_{\small y}(w,v)\rangle }{\partial\langle w,v\rangle}\right\rVert$$

For this particular problem, the solution is to use this with $g_{\small x}(w,v)=v$, and $g_{\small y}(w,v)=w-v$.

The rational is that we have $W=X+Y$ and a general transformation of pdf between coordinate systems when we have a bijection between $\langle X,Y\rangle$ and $\langle W, V\rangle$. So we will want some convenient $V$ that gives us such a transformation. Conveniently, $V:=X$ does so with minimal fuss.

$$\begin{align}f_{\small W,V}(w,v)&=f_{\small X,Y}(v,w-v)\cdot\left\lVert\dfrac{\partial\langle v,w-v\rangle }{\partial\langle w,v\rangle}\right\rVert\\&=f_{\small X,Y}(v,w-v)\cdot 1\end{align}$$

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Of course, we could just apply the transformation between, $\langle X,Y\rangle$ and $\langle W, X\rangle$

$$\begin{align}f_{\small W,X}(w,x)&=f_{\small X,Y}(x,w-x)\cdot\left\lVert\dfrac{\partial\langle x,w-x\rangle }{\partial\langle w,x\rangle}\right\rVert\\&=f_{\small X,Y}(x,w-x)\cdot 1\end{align}$$

Answer 2

Here is what I find to be an easier approach to this problem. Note that $$f_{X,Y}(x,y) = 4e^{-2(x+y)}\cdot\mathsf 1_{(0,\infty)\times(0,\infty)}(x,y) = \left(2e^{-2x}\cdot\mathsf 1_{(0,\infty)}(x) \right) \left(2e^{-2y}\cdot\mathsf 1_{(0,\infty)}(y) \right), $$ so $f_{X,Y} = f_Xf_Y$ and $X,Y$ are independent. Therefore $X$ and $Y$ have exponential distribution with rate $2$. We can find the density of $W=X+Y$ by convolution:

\begin{align} f_W(w) &= (f_X\star f_Y)(w)\\ &= \int_{\mathbb R} f_X(z)f_Y(w-z)\ \mathsf dz\\ &= \int_0^w 2e^{-2z}\cdot 2e^{-2(w-z)}\ \mathsf dz\\ &= 4e^{-2w} \int_0^w \mathsf dz\\ &= 4we^{-2w}\cdot\mathsf 1_{(0,\infty)}(w). \end{align}

Answer 3

I think you make the problem more complicated that it is. From statement you have $Y=W-X$, which gives: $$ J= \begin{bmatrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial x} & \frac{\partial y}{\partial w} \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \\ \end{bmatrix} $$

Step $X=V$ put here to make solution look more familiar.