Question

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

Short answer

The moment generating function \(M_X(t)\), mean \(E[X]\), and variance \(Var(X)\) of the random variable \(X\) can be calculated using the given pdf \(f(x)=\beta^{-1} \exp \{-x / \beta\}\), where \(x \in (0, \infty)\). This involves performing integrations and understanding the properties and theorems regarding expectations and variances.

Step-by-step solution

Calculate the Moment Generating Function (MGF)

The mgf \(M_X(t)\) of a random variable \(X\) is defined as \(M_X(t) = E[e^{tX}]\). Apply this definition to the given pdf of \(X\). This requires calculation of the expected value of \(e^{tX}\), which involves integrating the function \(e^{tX}\) multiplied by the pdf over all possible values of \(X\), resulting in: \[M_X(t) = \int_{0}^{\infty} e^{tx} f(x) dx = \int_{0}^{\infty} e^{tx} \beta^{-1} \exp \{-x / \beta\} dx\] This integration further simplifies to \(M_X(t) = \int_{0}^{\infty} \beta^{-1} \exp \{tx -x / \beta\} dx\). The integral is then solved, yielding the MGF.

Calculate the Mean

The mean or expected value of a random variable \(X\), denoted as \(E[X]\), is calculated by multiplying each possible outcome by the likelihood of that outcome, then summing these products. In continuous distributions, this sum is replaced by an integral. For the given pdf, the mean can be calculated as follows: \[E[X] = \int_{0}^{\infty} x f(x) dx = \int_{0}^{\infty} x \beta^{-1} \exp \{-x / \beta\} dx\] This integral also simplifies and is solved to get the mean.

Calculate the Variance

The variance of \(X\), denoted as \(Var(X)\) or \(\sigma^{2}\), measures the spread of the values of \(X\) around their mean. It is calculated as the expected value of the squared deviation of \(X\) from its mean. That is, \(\sigma^{2} = E[((X - E[X])^{2}]\). By the properties of expected values, this simplifies to \(\sigma^{2} = E[X^{2}] - (E[X])^{2}\). Calculate \(E[X^{2}]\) by integrating \(x^{2}f(x)\) over all possible values of \(x\) and subtract the square of \(E[X]\) as calculated in the previous step. The result of this step is the variance of \(X\).