# distribution of sum of samples of a random variable

I have a random variable X (which has some complicated PDF, it could be approximated to exponential), and I'm trying to find the distribution of another random variable Y, which is the sum of j samples of (1/X)

$Y_j=\sum_{i=1}^j(1/Xi)$

is there a way to find the distribution, CDF or CCDF of Y? regards

Following are the steps to be followed,

1. Find the distribution of $\frac{1}{X}$ by using the transformation $Z=\frac{1}{X} \implies \left|\frac{\partial X}{\partial Z}\right| = \frac{1}{Z^2}$

$$f_{Z}(z) = f_{X}\left(\frac{1}{z}\right)\frac{1}{z^2}$$

1. Find the moment generating function of $Z$ which is $M_Z(t) = E(e^{tZ})$.

2. Compute the moment generating function of $Y_j = \sum\limits_{i=1}^{j}Z_i$ where $Z_i$'s are iid random variables following the distribution $f_Z(z)$.

$$M_{Y_j}(t) = E(e^{tY_j}) = E(e^{t\sum\limits_{i=1}^{j}Z_i}) = \prod\limits_{i=1}^{j}E(e^{tZ_i}) = \prod\limits_{i=1}^{j}M_{Z_i}(t) = M_{Z}(t)^j$$

Note that in $3$ I used the independence of $Z_i$'s in third equality and $M_{Z_i}(t) = M_{Z}(t)$ in last equality since both $Z_i$ and $Z$ have same distribution.

Alternative $2$ and $3$ steps if the moment generating function does not correspond to commonly known distributions,

1. Let $U_1 = Y_j = \sum\limits_{i=1}^{j}Z_i$ and $U_i=Z_i, \forall i \in \left\{2,3,\ldots j \right\}$, then $Z_1 = U_1 - \sum\limits_{i=2}^{j}U_i$ and $Z_i=U_i, \forall i \in \left\{2,3,\ldots j \right\}$. The determinant of Jacobian of $Z$ with respect to $U$ will be equal to $1$. Then,

$$f_{U_1, U_2, \ldots, U_j}(u_1, u_2, \ldots, u_j) = f_{Z_1, Z_2, \ldots, Z_j}(u_1-\sum\limits_{i=2}^{j}u_i, u_2, \ldots, u_j) \cdot |det(J)|$$

$$f_{U_1, U_2, \ldots, U_j}(u_1, u_2, \ldots, u_j) = f_{Z_1}(u_1-\sum\limits_{i=2}^{j}u_i)f_{Z_2}(u_2)\ldots f_{ Z_j}(u_j) \cdot 1$$

The above equation follows from the independence of $Z_i$'s.

1. Integrate the joint pdf with respect to variables $U_2, U_3, \ldots, U_j$ to get the pdf of $U_1 (=Y_j)$.

$$\int_{u_1, u_2, \ldots u_j}f_{U_1, U_2, \ldots, U_j}(u_1, u_2, \ldots, u_j) \partial u_1 \partial u_2 \ldots \partial u_j = \int_{u_1, u_2, \ldots u_j}f_{Z_1}(u_1-\sum\limits_{i=2}^{j}u_i)f_{Z_2}(u_2)\ldots f_{ Z_j}(u_j)\partial u_1 \partial u_2 \ldots \partial u_j$$

The resulting function in $u_1$ will be the pdf of $U_1 = Y_j$.