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I'll state what I'm trying to prove below.

For a Poisson process $N(t) \sim \operatorname{Poisson}(\lambda t)$, $$ P\left(S_n \leq t\right)=P\left(N(t) \geq n\right)=1-P(N(t)<n), $$ where $S_n=\sum_{i=1}^n X_i$ and $X_i \stackrel{i i d}{\sim} \operatorname{Exponential}(\lambda)$.

I'm trying to prove this by showing the Laplace-Stieltjes transform of both sides are equivalent, where for a random variable $Y$ with distribution function $F_Y(y)$, the transform is defined as $$ \mathcal{L}\{F_Y(y)\}(s)=\int_0^{\infty} e^{-s y} d F_Y(y). $$ I'm also using the fact that $\mathcal{L}\{F_Y(y)\}(s)=\frac{1}{s}M_Y(-s)$.

For fixed $n$, I know $M_{S_n}(s)=\left[\frac{\lambda}{\lambda-s}\right]^n\Rightarrow \mathcal{L}\{P(S_n\leq t)\}(s)=\frac{1}{s}\left[\frac{\lambda}{\lambda+s}\right]^n$

But then $\mathcal{L}\{1-P(N(t)<n)\}(s)=\frac{1}{s}-\frac{1}{s}\exp(\lambda t(e^{-s}-1))$

Is there a flaw in my understanding? These two quantities are seemingly quite different

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2 Answers 2

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To complete the proof using the method you propose, you need to derive the Laplace transform of the latter distribution function and show that it is equivalent to the Laplace transform of the distribution of $S_n$. To do this, define the CDF:

$$H_n(t) \equiv \mathbb{P}(N(t) \geqslant n) = \sum_{k=n}^\infty \text{Pois}(k| \lambda t) = \sum_{k=n}^\infty \frac{(\lambda t)^k}{k!} e^{-\lambda t},$$

with the corresponding density function:

$$\begin{align} h_n(t) &= \frac{d H_n}{dt}(t) = \lambda \sum_{k=n}^\infty (k - \lambda t) \frac{(\lambda t)^{k-1}}{k!} e^{-\lambda t}. \\[6pt] \end{align}$$

We then have:

$$\begin{align} \mathcal{L}\{H_n\}(s) &= \int \limits_0^{\infty} e^{-st} d H_n(t) \\[6pt] &= \int \limits_0^{\infty} e^{-st} h_n(t) \ dt \\[6pt] &= \lambda \sum_{k=n}^\infty \int \limits_0^{\infty} (k - \lambda t) \frac{(\lambda t)^{k-1}}{k!} e^{-(\lambda+s) t} \ dt \\[6pt] &= \lambda \sum_{k=n}^\infty \bigg[ \int \limits_0^{\infty} \frac{(\lambda t)^{k-1}}{(k-1)!} e^{-(\lambda+s) t} \ dt - \int \limits_0^{\infty} \frac{(\lambda t)^{k}}{k!} e^{-(\lambda+s) t} \ dt \bigg] \\[6pt] &= \sum_{k=n}^\infty \bigg[ \Big( \frac{\lambda}{\lambda+s} \Big)^{k} \int \limits_0^{\infty} \text{Ga}(t|k,\lambda+s) \ dt - \Big( \frac{\lambda}{\lambda+s} \Big)^{k+1} \int \limits_0^{\infty} \text{Ga}(t|k+1,\lambda+s) \ dt \bigg] \\[6pt] &= \sum_{k=n}^\infty \bigg[ \Big( \frac{\lambda}{\lambda+s} \Big)^{k} - \Big( \frac{\lambda}{\lambda+s} \Big)^{k+1} \bigg] \\[12pt] &= \Big( \frac{\lambda}{\lambda+s} \Big)^{n} - \Big( \frac{\lambda}{\lambda+s} \Big)^{n+1} + \Big( \frac{\lambda}{\lambda+s} \Big)^{n+1} - \Big( \frac{\lambda}{\lambda+s} \Big)^{n+2} + \Big( \frac{\lambda}{\lambda+s} \Big)^{n+2} - \cdots \\[12pt] &= \Big( \frac{\lambda}{\lambda+s} \Big)^{n}. \\[6pt] \end{align}$$

(In the above working we use the shape/rate parameterisation of the gamma distribution.) This establishes that $\mathcal{L}\{H_n\}(s) = \mathcal{L}\{S_n\}(s)$, which establishes the distributional equivalence at issue.

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  • $\begingroup$ I see. Actually my motivation for asking comes from a more general problem, where I'm using a mixture of exponentials model for the inter-arrival times, and thus I didn't have immediate access to the CDF or density of this sum. I was hoping I could skirt the issue by using transform identities, but that might not be possible $\endgroup$
    – statian
    Sep 14, 2023 at 2:26
  • $\begingroup$ That is fine, and that's exactly what you've accomplished here. In the above working we show equivalence of the Laplace transforms so we don't need to derive the CDF of $S_n$ as part of the working --- we just use the known form for an exponential distribution and the rules of convolutions on the Laplace transform. $\endgroup$
    – Ben
    Sep 14, 2023 at 5:23
  • $\begingroup$ But I still do need the form of $P(N(t)\geq n)$ though correct? Meaning there's no way to use transform properties to compute $\mathcal L\{P(N(t)\geq n)\}(s)$ without using the density? $\endgroup$
    – statian
    Sep 14, 2023 at 12:50
  • $\begingroup$ Yes, you would probably still use that (though mostly for convenience). Re the use of the corresponding density of $H_n$, note that the Laplace transform is well-defined by the Lebesgue-Stieltjes integral (taken with respect to $H_n$, not $h_n$) so it is already well-defined without referring to the density. The easiest way to compute the integral is with the density, and it is simple to compute the density through differentiation of $H_n$. $\endgroup$
    – Ben
    Sep 14, 2023 at 22:14
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This answer proceeds by assuming that you're not explicitly required to use transforms. Then, this is solvable by examining the requisite CDFs. First, note that the sum $S_n$ of $n$ IID exponential random variables of the form $X_i \overset{i.i.d.}{\sim} \text{Exp}(\lambda)$ is Gamma distributed as $S_n \sim \text{Gamma}(n, \lambda)$ (using the shape + rate characterization of the gamma distribution). This can be readily verified by examining the product of MGFs for the $X_i$ and comparing it to the MGF for a gamma distribution.

We now evaluate the CDF of $S_n$ at $t$. In general, this would require us to evaluate $\gamma(n, \lambda t) / \Gamma(n)$, but since $n$ is an integer (i.e., $S_n$ is also Erlang-distributed), we can use the form

$$F_{S_n}(t) = P(S_n \leq t) = 1 - \sum_{i=0}^{n-1} \frac{(\lambda t)^i}{i!}\exp(-\lambda t),$$

which, by inspection, is equal to the complementary Poisson CDF for some random variable with distribution $\text{Poisson}(\lambda t)$; i.e., $N(t)$. This is exactly what you have on the rightmost portion of what you're trying to prove.

Intuitively, this checks out since $S_n$ is a random variable representing "at least $n$ occurrences of the event of interest before time $t$," which is a Poisson RV. Note that some similar questions have been addressed on Math StackExchange: 1, 2, 3.

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