Finding the joint PDF of transformations of two exponential RVs

Question

Given U and V which are exponential RVs with parameter $\lambda$, how might you find the joint PDF of X and Y where $X=\frac{U}{V}$ and $Y=U+V$.

I tried re-expressing U and V in terms of X and Y, because I've seen online examples where this is put into a matrix, and the subsequent determinant is used to get the joint PDF... how might I implement this or solve it by a different method.

After re-expression I got $U=\frac{Y}{1+1/X}$, and $V=\frac{Y}{1+X}$ but not sure how this would fit in the matrix.

Answer 1

Let $f_U$ and $f_V$ be the densities of $U$ and $V$. We compute the density of $Y=U+V$ by convolution: \begin{align} f_Y(t) &= f_U\star f_V(t)\\ &= \int_\mathbb R f_U(t-\tau)f_V(\tau)\ \mathsf d\tau\\ &= \int_0^t \lambda e^{-\lambda(t-\tau)}\lambda e^{-\lambda\tau}\ \mathsf d\tau\\ &= \lambda^2e^{-\lambda t}\int_0^t\ \mathsf d\tau\\ &= (\lambda t)\lambda e^{-\lambda t}. \end{align} We compute the density of the ratio $X=\frac UV$ by \begin{align} f_X(t) &= \int_\mathbb R |v|f_{U,V}(tv, v)\ \mathsf dv\\ &= \int_0^\infty v \lambda e^{-\lambda tv}\lambda e^{-\lambda v}\ \mathsf dv\\ &= \frac1{(1+t)^2}. \end{align}

Answer 2

Expanding on Math1000's answer, as to the derivation of the formulae.

For $Y$, we just use the definition of p. density function, and the fundamental theorem of calculus. (The p.d.f. is the unsigned derivative of the C.D.F..)

$$\begin{align}f_Y(y)&{}=\begin{vmatrix}\dfrac{\mathrm d ~~}{\mathrm d y}\mathsf P(U+V\leqslant y)\end{vmatrix} \\ &= \begin{vmatrix}\displaystyle\dfrac{\mathrm d ~~}{\mathrm d y}\int_\Bbb R\int_{-\infty}^{y-v} f_{U,V}(u,v)~\mathrm d u~\mathrm d v\end{vmatrix} \\&= \int_\Bbb R f_{U,V}(y-v, v)~\mathrm d v\end{align}$$

For $X$ we do the same, with complication that we have a ratio rather than a linear combination.   Now, since $U,V$ are exponential, then they are both almost certainly positive, and so therefore we are assured that $\mathsf P(U/V\leqslant x)=\mathsf P(U\leqslant xV)$.

$$\begin{align}f_X(x)&{}=\begin{vmatrix}\dfrac{\mathrm d ~~}{\mathrm d x}\mathsf P(U\leqslant xV)\end{vmatrix} \\ &= \begin{vmatrix}\displaystyle\dfrac{\mathrm d ~~}{\mathrm d x}\int_\Bbb R\int_{-\infty}^{xv} f_{U,V}(u,v)~\mathrm d u~\mathrm d v \end{vmatrix}\\&= \int_\Bbb R \begin{vmatrix}\dfrac{\partial xv}{\partial x}\end{vmatrix} f_{U,V}(xv, v)~\mathrm d v\\&= \int_\Bbb R \lvert v\rvert~f_{U,V}(xv, v)~\mathrm d v\end{align}$$

Everything else is just substituting the joint probability density function with the indicated arguments, then evaluating the intergrals.

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Now as to the joint distributeon of $X,Y$. As noted $X=U/V, Y=U+V$ so $U(X+1)=XY, V(X+1)=Y$

$$\begin{align}f_{X,Y}(x,y) {}&= \begin{vmatrix}\dfrac{\partial^2}{\partial x~\partial y}\mathsf P(U/V \leq x, U+V \leq y)\end{vmatrix} \\ &= \begin{Vmatrix}\dfrac{\partial [xy(x+1)^{-1}, y(x+1)^{-1}]}{\partial[x,y]}\end{Vmatrix} f_{U,V}(xy(x+1)^{-1}, y(x+1)^{-1}) \\ &= y(x+1)^{-2}\, f_{U,V}(xy(x+1)^{-1}, y(x+1)^{-1})\end{align}$$

Answer 3

Here are some hints:

first, note that the joint pdf of $(X, Y)$,i.e., $f(x,y)$ can be derived by differentiating their joint cdf: $$ f(x,y)=\dfrac {\partial ^2 F(x, y)}{\partial x \partial y} $$ so we need to find $$ \begin{align} F(x,y)&=\Pr \{X\leq x, Y\leq y\}\\ &=\Pr \{U/V\leq x,U+V\leq y\}\\ \end{align} $$

Assume $U,V$ are independent. Draw the lines of $U=xV$ and $U=y-V$ in the $(V,U)$ space and note that exponential RV has support $[0,\infty)$. Then there is a triangular area created by these three lines. Calculating the double integral would give us $F(x,y)$.