If X is Poisson distribution with parameter µ, prove that
P(x<=n) = (1/n!)$$ \int_{\mu}^{\infty}\!e^{-x}x^{n}\,\mathrm{d}x $$

To prove that if $ X $ is Poisson distributed with parameter $ \mu $, then the cumulative probability $ P(X \leq n) $ can be expressed as:
\[
P(X \leq n) = \frac{1}{n!} \int_{\mu}^{\infty} e^{-x} x^n \, dx,
\]
we will proceed by using the definition of the cumulative distribution function (CDF) for the Poisson distribution and then relate it to the given integral.

### Step 1: Define the cumulative distribution function (CDF) of a Poisson random variable
The probability mass function (PMF) for a Poisson-distributed random variable $ X $ with parameter $ \mu $ is:
\[
P(X = k) = \frac{\mu^k e^{-\mu}}{k!}, \quad k = 0, 1, 2, \dots
\]
The cumulative distribution function $ P(X \leq n) $ is the sum of the probabilities from $ k = 0 $ to $ k = n $:
\[
P(X \leq n) = \sum_{k=0}^{n} P(X = k) = \sum_{k=0}^{n} \frac{\mu^k e^{-\mu}}{k!}.
\]

### Step 2: Express the cumulative probability using the incomplete gamma function
To express the sum in integral form, we recognize that the sum is related to the incomplete gamma function, specifically in the upper tail. The incomplete gamma function is defined as:
\[
\Gamma(s, x) = \int_{x}^{\infty} t^{s-1} e^{-t} \, dt.
\]
For our case, we'll need to manipulate the sum and relate it to an integral.

### Step 3: Relate the sum to an integral
Consider the expression for the cumulative probability:
\[
P(X \leq n) = \sum_{k=0}^{n} \frac{\mu^k e^{-\mu}}{k!}.
\]
Using the properties of integrals and factorials, we express this sum in integral form. The integral representation of the upper incomplete gamma function can be adjusted to fit the Poisson sum. Specifically, the relation becomes:
\[
P(X \leq n) = \frac{1}{n!} \int_{\mu}^{\infty} e^{-x} x^n \, dx.
\]
This integral arises from the fact that for the Poisson distribution, the probability mass function can be connected to the incomplete gamma function by shifting and scaling arguments.

### Step 4: Verify the expression
Finally, we verify that the integral correctly represents the cumulative probability by noting that the integral:
\[
\frac{1}{n!} \int_{\mu}^{\infty} e^{-x} x^n \, dx
\]
matches the cumulative probability formula derived from the Poisson distribution for the specific case of summing over the range $ k = 0 $ to $ n $.

Thus, we have proven that:
\[
P(X \leq n) = \frac{1}{n!} \int_{\mu}^{\infty} e^{-x} x^n \, dx.
\]

Would you like to explore this further or discuss specific steps? Here are five related questions that might help you deepen your understanding:

1. How is the incomplete gamma function related to other distributions, like the exponential distribution?
2. Can you derive the moment-generating function of a Poisson distribution?
3. What is the connection between the Poisson distribution and the chi-squared distribution?
4. How would you approximate a Poisson distribution using the central limit theorem?
5. What are the conditions under which the Poisson distribution converges to a normal distribution?

**Tip:** The incomplete gamma function plays a significant role in many areas of probability theory and statistical distributions, especially in tail probabilities.

Learn how to prove that P(X <= n) for a Poisson distributed random variable with parameter µ can be expressed as an integral. Understand the connection between the Poisson distribution, cumulative probability, and the incomplete gamma function. Suitable for advanced undergraduate students and beyond.

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