Distribution of sum of exponential variables with different parameters

We have $k$ independent random variables with exponential distribution ($T_1, T_2, \ldots , T_k$), parameters of random variables are ($\lambda,\frac{\lambda}{2},\frac{\lambda}{3},\ldots,\frac{\lambda}{k}$), what is the distribution of new variable $T = T_1 + T_2 + \cdots + T_k$

• are they independent ? Dec 12 '16 at 18:55
• @Canardini Oh sorry I forgot it, yes they are. Dec 12 '16 at 18:55
• Sometimes when people write of "the exponential distribution with parameter $\alpha$" they mean $\displaystyle e^{-x/\alpha} \left(\frac{dx} \alpha\right) \text{ for } x\ge 0,$ so that $\alpha$ is the expected value, and sometimes they mean $\displaystyle e^{-\alpha x} (\alpha\,dx) \text{ for } x\ge0,$ so that $1/\alpha$ is the expected value. Which do you have in mind here? $\qquad$ Dec 12 '16 at 19:07
• @MichaelHardy Thanks for your comment, the second one so when I say $T_i$ has exponential distribution ($Exp(\frac{\lambda}{i})$) I mean $e^{-\frac{\lambda}{i}T_i}$ Dec 12 '16 at 19:10
• I do not see any pattern when using moment generating functions. Do you know that if the exponentials are identically distributed, the sum is the Gamma distribution? Dec 12 '16 at 19:18

you can use the main result here applied to the sequence $\lambda, \lambda/2,...,\lambda/k$, which for the general case states that

$$h_{X_1,.,X_n}(x) = \left[ \prod_{i=1}^n \lambda_{i} \right] \sum_{j=\ 1}^n \frac{e^{-\lambda_i x}}{\prod_{k \neq j}^n(\lambda_i-\lambda_j)}$$

• Has anybody found any information with regards to the CDF equivalent of this?
– jab
Jun 1 '19 at 23:03

Hint: You could use convolution to calculate the distribution of two independent variables:

Assume $X$ follows $f(x)$, $Y$ follows $g(y)$, then $Z=X+Y$ follows

$$f_Z(z)=\int _{-\infty}^{+\infty}f_X(z-y)f_Y(y) \, dy.$$

Following this logic, you just do a serial integrations, then you would get the result.

• Can you please tell me about doing serial integration for 3 variables when we have $W = X + Y + Z$? Dec 12 '16 at 19:05
• So you just first get distribution $X+Y$, then use this obtained distribution to do the integration with $Z$. Two steps. Dec 12 '16 at 19:07
• This answer is correct as far as it goes, but it doesn't attempt to say whether the particular pattern in the sequence of parameters leads to some pattern in the distribution of the sequence of sums. Dec 12 '16 at 19:12