The Moment Generating Function (MGF) of a random variable is defined as:

$M_X(t) = \mathbb{E}\left[e^{tX}\right]$

For an exponential distribution with rate parameter $\lambda$, the probability density function (PDF) is:

$f_X(x) = \lambda e^{-\lambda x}, \quad x \geq 0$

Step 1: Deriving the MGF

The MGF of $X$ is given by:

$M_X(t) = \mathbb{E}\left[e^{tX}\right] = \int_{0}^{\infty} e^{tx} f_X(x) \, dx$

Substituting the PDF of the exponential distribution:

$M_X(t) = \int_{0}^{\infty} e^{tx} \lambda e^{-\lambda x} \, dx$

This simplifies to:

$M_X(t) = \lambda \int_{0}^{\infty} e^{x(t - \lambda)} \, dx$

To evaluate the integral, notice that it is a standard form integral:

$\int_{0}^{\infty} e^{-ax} \, dx = \frac{1}{a} \quad \text{for} \, a > 0$

Here, $a = \lambda - t$, so:

$M_X(t) = \lambda \cdot \frac{1}{\lambda - t} \quad \text{for} \, t < \lambda$

Thus, the MGF of the exponential distribution is:

$M_X(t) = \frac{\lambda}{\lambda - t}, \quad \text{for} \, t < \lambda$

Step 2: Finding the Mean and Variance

Mean ($\mu$)

The mean of $X$ can be found by differentiating the MGF with respect to $t$ and then evaluating at $t = 0$:

$\mu = \mathbb{E}[X] = M_X'(0)$

Differentiate $M_X(t)$ with respect to $t$:

$M_X'(t) = \frac{d}{dt} \left(\frac{\lambda}{\lambda - t}\right)$

Applying the quotient rule:

$M_X'(t) = \frac{0(\lambda - t) - \lambda(-1)}{(\lambda - t)^2} = \frac{\lambda}{(\lambda - t)^2}$

Evaluate at $t = 0$:

$M_X'(0) = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}$

Thus, the mean of the exponential distribution is:

$\mu = \frac{1}{\lambda}$

Variance ($\sigma^2$)

The variance can be found using the second derivative of the MGF:

$\text{Var}(X) = M_X''(0) - (M_X'(0))^2$

First, differentiate $M_X'(t)$ to find $M_X''(t)$:

$M_X''(t) = \frac{d}{dt} \left(\frac{\lambda}{(\lambda - t)^2}\right)$

Using the chain rule:

$M_X''(t) = \frac{2\lambda}{(\lambda - t)^3}$

Evaluate at $t = 0$:

$M_X''(0) = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2}$

Finally, calculate the variance:

$\text{Var}(X) = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$

Summary

• MGF of the exponential distribution: $\displaystyle M_X(t) = \frac{\lambda}{\lambda - t}, \quad \text{for} \, t < \lambda$
• Mean: $\displaystyle \mu = \frac{1}{\lambda}$
• Variance: $\displaystyle \text{Var}(X) = \frac{1}{\lambda^2}$

Would you like more details on any part of this explanation?

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Here are 5 related questions you might be interested in:

1. How does the exponential distribution relate to the Poisson distribution?
2. What is the relationship between the exponential distribution and the gamma distribution?
3. How do you find the MGF of a Poisson distribution?
4. What is the significance of the memoryless property in the exponential distribution?
5. Can you derive the MGF for the normal distribution?

Tip: The MGF can be a powerful tool for deriving moments, such as mean and variance, especially when dealing with complex distributions.