To solve the given questions, we analyze the properties of the exponential distribution $X \sim \text{Exponential}(\lambda)$, where $\lambda > 0$ is the rate parameter.

(a) Find $M(t) = \mathbb{E}(e^{tX})$

The moment-generating function (MGF) $M(t)$ is defined as: $M(t) = \mathbb{E}(e^{tX}) = \int_{0}^{\infty} e^{tx} f_X(x) \, dx$ where $f_X(x) = \lambda e^{-\lambda x}$ is the probability density function (PDF) of the exponential distribution. Substituting $f_X(x)$ into the integral: $M(t) = \int_{0}^{\infty} e^{tx} \lambda e^{-\lambda x} \, dx$

Combine the exponents: $M(t) = \lambda \int_{0}^{\infty} e^{(t - \lambda)x} \, dx$

The integral converges if $t < \lambda$. For $t < \lambda$, the integral evaluates as: \int_{0}^{\infty} e^{(t - \lambda)x} \, dx = \frac{1}{\lambda - t}, \quad \text{(valid for t < \lambda)}.

Thus: $M(t) = \frac{\lambda}{\lambda - t}, \quad t < \lambda.$

(b) Use the result obtained in (a) to find $\mathbb{E}(X)$

The expected value $\mathbb{E}(X)$ can be found from the first derivative of $M(t)$ at $t = 0$: $\mathbb{E}(X) = M'(0).$

From $M(t) = \frac{\lambda}{\lambda - t}$, calculate the derivative: $M'(t) = \frac{d}{dt} \left( \frac{\lambda}{\lambda - t} \right) = \frac{\lambda}{(\lambda - t)^2}.$

Substitute $t = 0$: $M'(0) = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}.$

Thus: $\mathbb{E}(X) = \frac{1}{\lambda}.$

(c) Use the result obtained in (a) to find $\text{Var}(X)$

The variance $\text{Var}(X)$ is given by: $\text{Var}(X) = \mathbb{E}(X^2) - [\mathbb{E}(X)]^2.$

First, find $\mathbb{E}(X^2)$ using the second derivative of $M(t)$ at $t = 0$: $\mathbb{E}(X^2) = M''(0).$

From $M'(t) = \frac{\lambda}{(\lambda - t)^2}$, calculate the second derivative: $M''(t) = \frac{d}{dt} \left( \frac{\lambda}{(\lambda - t)^2} \right) = \frac{2\lambda}{(\lambda - t)^3}.$

Substitute $t = 0$: $M''(0) = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2}.$

Thus: $\mathbb{E}(X^2) = \frac{2}{\lambda^2}.$

Now calculate $\text{Var}(X)$: $\text{Var}(X) = \mathbb{E}(X^2) - [\mathbb{E}(X)]^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2.$

Simplify: $\text{Var}(X) = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}.$

Final Results:

1. $M(t) = \frac{\lambda}{\lambda - t}, \quad t < \lambda.$
2. $\mathbb{E}(X) = \frac{1}{\lambda}.$
3. $\text{Var}(X) = \frac{1}{\lambda^2}.$

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Would you like a step-by-step breakdown or further clarification on any part of the solution?

Here are 5 related questions to deepen your understanding:

1. What happens to $M(t)$ when $t \geq \lambda$?
2. How is the MGF used to derive higher moments of a distribution?
3. What is the relationship between the rate parameter $\lambda$ and the mean of the exponential distribution?
4. Can you generalize the variance result for a transformed exponential variable $Y = aX$?
5. How does the exponential distribution relate to the Poisson process?

Tip: The exponential distribution is memoryless, meaning $P(X > x + t \mid X > t) = P(X > x)$, which simplifies many calculations!