; n n p t ∞

It looks like you are referring to a sum involving probabilities and a Poisson distribution, but the notation is incomplete. Let’s clarify the context and provide a more complete explanation.

Context: Poisson Distribution and Infinite Series

In the context of the Poisson distribution, we often deal with sums of probabilities over all possible numbers of events. Specifically, for a Poisson process with rate $\lambda$, the probability of observing $n$ events in a time interval $t$ is given by:

$P_n(t) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}$

When you sum these probabilities over all possible values of $n$ (from 0 to $\infty$), you are essentially considering the total probability, which must sum to 1.

Sum of Poisson Probabilities

To show that the sum of all probabilities in a Poisson distribution equals 1, we write:

$\sum_{n=0}^{\infty} P_n(t) = \sum_{n=0}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!}$

Proof Using Series Expansion

Let's prove this by using the series expansion of the exponential function:

1. Exponential Function Expansion: The exponential function $e^x$ can be expanded as a power series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

2. Applying the Expansion to Our Problem: In our case, we substitute $x = \lambda t$: $e^{\lambda t} = \sum_{n=0}^{\infty} \frac{(\lambda t)^n}{n!}$

3. Including the Exponential Term: Notice that $P_n(t)$ includes $e^{-\lambda t}$: $\sum_{n=0}^{\infty} P_n(t) = \sum_{n=0}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!}$ We can factor out $e^{-\lambda t}$ from the sum: $\sum_{n=0}^{\infty} \frac{(\lambda t)^n e^{-\lambda t}}{n!} = e^{-\lambda t} \sum_{n=0}^{\infty} \frac{(\lambda t)^n}{n!}$

4. Using the Series Expansion: From the exponential series expansion, we know: $\sum_{n=0}^{\infty} \frac{(\lambda t)^n}{n!} = e^{\lambda t}$

5. Simplifying: Substitute back into our sum: $e^{-\lambda t} \cdot e^{\lambda t} = 1$

Therefore, we have shown that the sum of the probabilities over all possible numbers of events in a Poisson process is 1, confirming that the Poisson distribution is a valid probability distribution.

Conclusion

This result confirms the normalization of the Poisson distribution, ensuring that the total probability is 1.

Do you want details or have any questions? Here are five questions you might ask next:

1. How do we calculate the mean and variance of a Poisson distribution?
2. Can you explain the derivation of the Poisson distribution from a binomial distribution?
3. What are some applications of the Poisson distribution in real life?
4. How does the Poisson distribution relate to the exponential distribution?
5. What is the probability generating function for the Poisson distribution?

Tip: When working with series and sums in probability, understanding the underlying exponential functions and their properties can greatly simplify the derivation and proof of important results.