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Suppose $c>0$. Let $X_i$ be independently and identically distributed exponential variables with parameter $\lambda >0$ and $$S_n=\sum_{i=1}^n X_i.$$ Let $I=\min \{i: X_i >c\}$. How do I show that $\mathbb{E} [S_{I-1}+c] = \dfrac{e^{\lambda c} -1}{\lambda}$?

I started as:

$S_{I-1}$ follows a Gamma distribution with parameters $(I-1,\lambda)$, where if a random variable $X$ follows a Gamma distribution with parameters $n$ and rate $\lambda$, then $$f(x)= e^{-\lambda x}\dfrac{{\lambda}^n x^{n-1}}{(n-1)!}.$$

Then I conditioned as follows:

$$\mathbb{E}[S_{I-1}]=\sum_{i=1}^{\infty} \mathbb{E}[S_{I-1}\mid I=i]\mathbb{P}(I=i).$$

Here I'm wondering whether I should sum from $i=2$ instead!

Anyway, I obtained $\mathbb{P}(I=i)=(1-e^{-\lambda c})^{i-1}e^{-\lambda c}$ as $$\mathbb{P}(I=i)=\mathbb{P}(X_1<c, X_2<c, \dots, X_{i-1}<c, X_i>c).$$

I have doubts here also because the way I did it I obtained $\mathbb{E}[S_{I-1}]=\dfrac{e^{\lambda c} -1}{\lambda}$, which is wrong.

Martin Sleziak
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Bhavish Suarez
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  • you want wald's identity ,http://en.wikipedia.org/wiki/Wald's_equation – mike May 24 '13 at 19:23
  • Be careful about the assertion that $S_{I-1}$ has a Gamma $(I-1,\lambda)$ distribution because it is the sum of $I-1$ independent exponential random variables with common parameter $\lambda$ random variables. – Dilip Sarwate May 24 '13 at 19:40

1 Answers1

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You are correct in your assertion that $I$ is a geometric random variable with parameter $e^{-\lambda c}$ and hence expected value $e^{\lambda c}$. But, given the value of $I$, all the $X_i, 1 \leq i < I$, are known to have value $\leq c$, and therefore each such $X_i$ has conditional marginal density function $$f_{X_i \mid I}(x) = \begin{cases}\frac{\lambda e^{-\lambda x}}{1-e^{-\lambda c}}, &0 \leq x \leq c,\\\quad\\ 0,&\text{otherwise}\end{cases}$$ which is just the truncation (and normalization) of the exponential density to the interval $[0,c]$. Hence, the conditional mean is $$\begin{align} E[X_i\mid I] &= \int_0^c \frac{x\lambda e^{-\lambda x}}{1-e^{-\lambda c}}\,\mathrm dx = \frac{1}{1-e^{-\lambda c}}\int_0^cx\lambda e^{-\lambda x} \,\mathrm dx\\ &= \left.\left.\frac{1}{1-e^{-\lambda c}}\right[-xe^{-\lambda x}\biggr|_0^c + \int_0^c e^{-\lambda x} \,\mathrm dx\right]\\ &= \left.\left.\frac{1}{1-e^{-\lambda c}}\right[-ce^{-\lambda c} + \left.\left.\frac{1}{\lambda}\right(1- e^{-\lambda c} \right)\right]\\ &= \frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1} \end{align}$$ which gives $$E[S_{I-1}\mid I] = E\left[\sum_{i=1}^{I-1}X_i\right] = (I-1)\left[\frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1}\right]$$ and $$E[S_{I-1}] = E\left[E[S_{I-1}\mid I]\right] = (e^{\lambda c}-1)\left[\frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1}\right] = \frac{e^{\lambda c}-1}{\lambda} - c.$$

Dilip Sarwate
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  • Thanks a lot! Say, why can't I assume that $S_{I-1}$ has a Gamma distribution? – Bhavish Suarez May 25 '13 at 06:26
  • I'm not able to derive the conditional density function. – Bhavish Suarez May 25 '13 at 10:00
  • You cannot assume that $S_{I-1}$ is a Gamma random variable because it is not the sum of $I-1$ independent exponential random variables, but rather of $I-1$ truncated exponential random variables. A Gamma random variable can take on all values in $(0,\infty)$ whereas, given $I$, $S_{I-1}$ has maximum value $c(I-1)$. As for the conditional density of $X_i$, note that this answer of André Nicolas to a different question of yours finds the conditional distribution of $X_i$ (he calls it $Y$), so take the derivative to get the density I used. – Dilip Sarwate May 25 '13 at 11:12
  • Oh okay..I got it, thanks! Actually I was trying to find the joint density function and I was stuck there. – Bhavish Suarez May 26 '13 at 08:13