Question

use porgeni (distribution) to show that e(x) = 1/λ \
var(x) = 1/λ

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, but the given expressions seem to have a typo. If we assume the random variable is \( X \) following Poisson distribution with parameter \( \lambda \), then:

Step2: Verify Expected Value

The probability mass function of Poisson distribution is \( P(X = k)=\frac{e^{-\lambda}\lambda^{k}}{k!}, k = 0,1,2,\dots \). The expected value \( E(X)=\sum_{k = 0}^{\infty}k\cdot\frac{e^{-\lambda}\lambda^{k}}{k!}=\lambda \). Wait, the given \( E(x)=\frac{1}{x} \) (maybe a typo, should be \( E(X)=\lambda \)) and \( Var(X)=\frac{1}{\lambda} \)? No, variance of Poisson is \( \lambda \). Wait, maybe it's an exponential distribution? For exponential distribution with parameter \( \lambda \) (rate parameter), the probability density function is \( f(x)=\lambda e^{-\lambda x}, x\geq0 \). Then:

Step3: Calculate Expected Value for Exponential

\( E(X)=\int_{0}^{\infty}x\cdot\lambda e^{-\lambda x}dx \). Using integration by parts (let \( u = x, dv=\lambda e^{-\lambda x}dx \), then \( du = dx, v=-e^{-\lambda x} \)), we get \( E(X)=-x e^{-\lambda x}\big|_{0}^{\infty}+\int_{0}^{\infty}e^{-\lambda x}dx = 0+\frac{1}{\lambda} \).

Step4: Calculate Variance for Exponential

First, \( E(X^{2})=\int_{0}^{\infty}x^{2}\cdot\lambda e^{-\lambda x}dx \). Using integration by parts twice, we find \( E(X^{2})=\frac{2}{\lambda^{2}} \). Then \( Var(X)=E(X^{2})-(E(X))^{2}=\frac{2}{\lambda^{2}}-\frac{1}{\lambda^{2}}=\frac{1}{\lambda^{2}} \)? Wait, no, that's not matching. Wait, maybe the parameter is \( \mu=\frac{1}{\lambda} \) (scale parameter). For exponential distribution with scale parameter \( \mu \), \( f(x)=\frac{1}{\mu}e^{-\frac{x}{\mu}}, x\geq0 \). Then \( E(X)=\mu \), \( Var(X)=\mu^{2} \). Wait, the given \( E(x)=\frac{1}{x} \) is confusing. Maybe the original problem has a typo, and it's about exponential distribution where \( E(X)=\frac{1}{\lambda} \) and \( Var(X)=\frac{1}{\lambda^{2}} \)? Or maybe Poisson with a miswritten parameter. Alternatively, if we consider the given as \( E(X)=\frac{1}{\lambda} \) and \( Var(X)=\frac{1}{\lambda} \), that would be a different distribution, but Poisson has \( E(X)=Var(X)=\lambda \), exponential has \( E(X)=\frac{1}{\lambda}, Var(X)=\frac{1}{\lambda^{2}} \) (rate parameter) or \( E(X)=\mu, Var(X)=\mu^{2} \) (scale parameter). Maybe the problem is to show for a certain distribution (maybe a typo in the problem statement) that mean and variance are \( \frac{1}{\lambda} \), but likely it's a miswritten Poisson or exponential. Assuming the intended is exponential with rate \( \lambda \), \( E(X)=\frac{1}{\lambda} \), and if we consider a different approach, but the key is that the problem likely involves probability distributions (Natural Science - subfield: Probability/Statistics) to show the mean and variance.

Answer:

Assuming the problem is about the exponential distribution with rate parameter \( \lambda \), \( E(X)=\frac{1}{\lambda} \) and (if there's a typo and variance is \( \frac{1}{\lambda^{2}} \), but the given has \( Var(X)=\frac{1}{\lambda} \), maybe a different distribution). However, the main subfield here is Statistics (under Mathematics) or Probability (under Mathematics) to analyze the expected value and variance of a random variable. The correct derivation (for exponential with rate \( \lambda \)) shows \( E(X)=\frac{1}{\lambda} \) via integration, and if we correct the variance, but based on the given (maybe a typo), the subfield is Statistics (Mathematics) to handle the expectation and variance calculations.