I have two i.i.d random variables $X_{1}$ and $X_{2}$ following a continuous Poisson distribution function
$P(x) = \lambda e^{-\lambda\cdot x}$.
I wish to obtain a distribution function of sum of two i.i.d random variables;
$Y = Y_{1}+Y_{2}$, where,
$Y_1 = \frac{1}{X_{1}} , Y_2 = \frac{1}{X_{2}}$.
To do this I would guess two steps-
1) Obtain the inverse Poisson distribution g(y) for $Y=\frac{1}{X}$
2) Obtain the distribution for sum of random variables $Y_{1}+Y_{2}$
I need help on how should I approach this problem?
Best Answer
Hint: If $X\sim Exp(\lambda)$, its cdf is $F_X(x)=1-e^{-\lambda x}$, $x\geq 0$. Let us deduce the cdf of $1/X$:
$P(1/X\leq x)=P(X\geq 1/x)=1-P(X\leq 1/x)=1-(1-e^{-\lambda/x})=e^{-\lambda/x}$, $x>0$. Its pdf is $g(x)=\displaystyle\frac{\lambda}{x^2}e^{-\lambda x}$, $x>0$.
If $X_1, X_2$ are i.i.d. then $1/X_1, 1/X_2$ are also i.i.d., because: $P(1/X_1\leq x_1, 1/X_2\leq x_2)=P(X_1\geq 1/x_1, X_2\geq 1/x_2)=P(X_1\geq 1/x_1)P(X_2\geq 1/x_2)= P(1/X_1\leq x_1, 1/X_2\leq x_2)$.
Thus the sum $Y=Y_1+Y_2$ has as pdf the convolution of the pdf of $Y_1$ and $Y_2$, i.e. denoting by $g_1, g_2$ the pdfs of $Y_1, Y_2$ we have $h(z)$ as pdf for $Y$, where $h(z)=\int_\epsilon^\infty g_1(z-y)g_2(y)dy$, where $\epsilon$ is positive and close to zero.