Sum of i.i.d. random variables I am trying to obtain the distribution of $Y = \sum_{i=1}^{N} X_{i}$ when the distribution of each i.i.d. random variable $X_{i}$ is given by
\begin{align}
f_{X_{i}}(x) = \begin{cases}\lambda e^{\lambda x}, & 0 \le x \le R\\ 0, & \text{otherwise}\end{cases}
\end{align}
where $R = \ln 2 / \lambda$ so $\int_{-\infty}^{\infty} f_{X_{i}}(x) = 1$.
You can easily notice that the random variable $X_{i}$ is distributed as horizontally flipped from the exponential distribution.
I am in the need to find what distribution $Y$ follows. Can anybody help?
 A: Hope this helps. We can easily find a closed form of the characteristic function. Find the CF of the rv $X_1$:
$$\begin{aligned}E[e^{i\theta X_1}]&=\lambda\int_\mathbb{R}\mathbb{I}_{[0,R]}e^{\lambda x}e^{ix\theta}dx=\\
&=\lambda \int_{[0,R]}e^{x(\lambda +i\theta)}dx=\\
&=\lambda \frac{e^{R(\lambda + i\theta)}-1}{\lambda + i \theta}\end{aligned}$$
By IIDness:
$$E[e^{i\theta Y}]=E[e^{i\theta X_1}]^n=\lambda^n\bigg(\frac{e^{R\lambda}e^{i\theta R}-1}{\lambda+i\theta}\bigg)^n=\lambda^n\bigg(\frac{2e^{i\theta R}-1}{\lambda+i\theta}\bigg)^n$$
With numerical integration we can see the shape of the density of $Y_n$ and how it changes with $n$ (as expected, it becomes more and more bell shaped):
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A: If you set $Y_i = R-X_i$, you find that $Y_i$ has the truncated exponential distribution. There are some papers on sums of truncated exponential distributions, for example Nath (1974). Check also the CrossValidated post "PDF of sum of truncated exponential distribution".
Nath, G. Baikunth, On sums and products of truncated exponential variates, Scand. Actuarial J. 1974, 205-210 (1974). ZBL0292.62015.
