use porgens theorem to show that
e(x) = 1/λ
var(x) = 1/λ

Question

Accepted Answer

# Explanation:
## Step1: Recall Poisson Distribution Properties
For a Poisson random variable $ X $ with parameter $ \lambda $, the expected value (mean) $ E(X) $ and variance $ Var(X) $ are both equal to $ \lambda $? Wait, no, wait—wait, maybe there's a typo or miswriting. Wait, actually, for Poisson distribution, $ E(X)=\lambda $ and $ Var(X)=\lambda $. But the problem here has $ E(X)=\frac{1}{\lambda} $ and $ Var(X)=\frac{1}{\lambda} $? Wait, maybe it's an exponential distribution? Wait, no, exponential distribution has $ E(X)=\frac{1}{\lambda} $ and $ Var(X)=\frac{1}{\lambda^2} $. Wait, maybe the problem is about a Poisson distribution with rate $ \mu=\frac{1}{\lambda} $? Let's re - examine.

Wait, the user wrote "use poisson distribution to show that $ E(X)=\frac{1}{x} $" (probably $ \frac{1}{\lambda} $) and $ Var(X)=\frac{1}{\lambda} $. Wait, maybe there's a mix - up. Let's start with the Poisson distribution. The probability mass function of a Poisson random variable $ X $ is $ P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!} $, for $ k = 0,1,2,\cdots $, where $ \lambda>0 $ is the parameter.

### Step 1: Calculate the Expected Value $ E(X) $
The expected value of a discrete random variable $ X $ is given by $ E(X)=\sum_{k = 0}^{\infty}k\cdot P(X = k) $.

For Poisson distribution:
$$
\begin{align*}
E(X)&=\sum_{k = 0}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\
&=e^{-\lambda}\sum_{k = 1}^{\infty}\frac{\lambda^{k}}{(k - 1)!}\
&=\lambda e^{-\lambda}\sum_{m = 0}^{\infty}\frac{\lambda^{m}}{m!}\quad(\text{let }m=k - 1)\
&=\lambda e^{-\lambda}e^{\lambda}\quad(\text{since }\sum_{m = 0}^{\infty}\frac{\lambda^{m}}{m!}=e^{\lambda})\
&=\lambda
\end{align*}
$$

### Step 2: Calculate the Variance $ Var(X) $
The variance $ Var(X)=E(X^{2})-[E(X)]^{2} $. First, find $ E(X^{2}) $.

We know that $ E[X(X - 1)]=E(X^{2})-E(X) $.

$$
\begin{align*}
E[X(X - 1)]&=\sum_{k = 0}^{\infty}k(k - 1)\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}\
&=e^{-\lambda}\sum_{k = 2}^{\infty}\frac{\lambda^{k}}{(k - 2)!}\
&=\lambda^{2}e^{-\lambda}\sum_{m = 0}^{\infty}\frac{\lambda^{m}}{m!}\quad(\text{let }m = k - 2)\
&=\lambda^{2}e^{-\lambda}e^{\lambda}\
&=\lambda^{2}
\end{align*}
$$

Then $ E(X^{2})=E[X(X - 1)]+E(X)=\lambda^{2}+\lambda $.

And $ [E(X)]^{2}=\lambda^{2} $. So $ Var(X)=(\lambda^{2}+\lambda)-\lambda^{2}=\lambda $.

Wait, but the problem states $ E(X)=\frac{1}{\lambda} $ and $ Var(X)=\frac{1}{\lambda} $. Maybe the parameter is $ \mu=\frac{1}{\lambda} $. Let's re - define the Poisson parameter as $ \mu=\frac{1}{\lambda} $. Then $ E(X)=\mu=\frac{1}{\lambda} $ and $ Var(X)=\mu=\frac{1}{\lambda} $. That must be the case. So if we let the Poisson parameter be $ \mu=\frac{1}{\lambda} $, then:

For $ E(X) $:
$$
E(X)=\mu=\frac{1}{\lambda}
$$

For $ Var(X) $:
$$
Var(X)=\mu=\frac{1}{\lambda}
$$

So by using the Poisson distribution with parameter $ \mu=\frac{1}{\lambda} $, we can show that the expected value and variance are both $ \frac{1}{\lambda} $.

# Answer:
To show $ E(X)=\frac{1}{\lambda} $ and $ Var(X)=\frac{1}{\lambda} $ using Poisson distribution, we consider a Poisson random variable $ X $ with parameter $ \mu=\frac{1}{\lambda} $.

### Step 1: Expected Value Calculation
The expected value of a Poisson random variable $ X $ with parameter $ \mu $ is $ E(X)=\mu $. Substituting $ \mu = \frac{1}{\lambda} $, we get $ E(X)=\frac{1}{\lambda} $.

### Step 2: Variance Calculation
The variance of a Poisson random variable $ X $ with parameter $ \mu $ is $ Var(X)=\mu $. Substituting $ \mu=\frac{1}{\lambda} $, we get $ Var(X)=\frac{1}{\lambda} $.

Thus, for a Poisson random variable with parameter $ \mu=\frac{1}{\lambda} $, $ E(X)=\frac{1}{\lambda} $ and $ Var(X)=\frac{1}{\lambda} $.

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Question

Explanation:

Step1: Recall Poisson Distribution Properties

Step 1: Calculate the Expected Value \( E(X) \)

Step 2: Calculate the Variance \( Var(X) \)

Answer:

Step 1: Expected Value Calculation

Step 2: Variance Calculation