Question

Let \(X\) have the exponential pdf, \(f(x)=\beta^{-1} \exp \\{-x / \beta\\}, 0

Short answer

The Moment Generating Function (MGF) is \(M(t) = 1 / (1 - t\beta)\) for \(t<1/\beta\). The mean is \(\beta\) and the variance is \(\beta^2\).

Step-by-step solution

Compute the Moment Generating Function (MGF)

The Moment Generating Function (MGF) of a random variable with a PDF f(x) is given by \(M(t) = E(e^{tx}) = \int_{-\infty}^{+\infty} e^{tx} \cdot f(x) \, dx\) . With \(f(x) = \beta^{-1}\exp \{-x / \beta\}\) we find \[M(t) = \int_{0}^{+\infty} e^{tx} \cdot \beta^{-1} \exp \{-x / \beta\} \, dx\] . This integral is computed by substituting \(x / \beta\) with a new variable \(z\) where \(dz = dx / \beta \). Now we get \[ M(t) = \int_{0}^{+\infty} e^{t\beta z} \cdot \exp \{-z\} \, dz \] Using integral properties, it is computed as \(M(t) = 1 / (1 - t\beta)\), for \(t<1/\beta\).

Compute the Mean

The mean of the random variable X, denoted as \(\mu\) or \(E[X]\), is obtained by differentiating the mgf and evaluating at \( t = 0\). As \(M'(t)=\mu=E[X]= \frac{d}{dt}M(t)=\frac{d}{dt}(1/(1-\beta t))\). Setting \( t = 0\) we get \(\mu = \beta\).

Compute the Variance

The variance of the random variable X, denoted as \(Var[X]\) or \(\sigma^2\), is obtained by differentiating the mgf twice and evaluating at \( t = 0\). As \(M''(t) =\sigma^2=E[X^2]-(E[X])^2= \frac{d^2}{dt^2}M(t)=\frac{d^2}{dt^2}(1/(1-\beta t))\). Setting \( t = 0\) we get \(\sigma^2 =\beta^2\).