Question

Let the random variable \(X\) have mean \(\mu\), standard deviation \(\sigma\), and \(\mathrm{mgf}\) \(M(t),-h

Short answer

The standard form of the random variable has an expectation of 0, variance of 1, and its moment generating function simplifies to \(e^{-\mu t / \sigma} M \left(\frac{t}{\sigma}\right), -h \sigma < t < h \sigma\)

Step-by-step solution

Compute Expectation of Standard Form

We have to compute \[E\left(\frac{X-\mu}{\sigma}\right)\] Remember, the expectation operator distributes over addition and scalar multiplication.\nTherefore, we can break this down and compute:\n\[\frac{1}{\sigma}E(X) - \frac{\mu}{\sigma}E(1)\] By definition, \(E(X) = \mu\) and \(E(1) = 1\), so this simplifies further to:\n\[\frac{\mu}{\sigma} - \frac{\mu}{\sigma} = 0\]

Compute Expectation of Squared Standard Form

Now we compute \[E\left[ \left(\frac{X-\mu}{\sigma}\right)^2 \right]\]. Using similar steps as before, this simplifies to \[\frac{1}{\sigma^2}E((X - \mu)^2)\] The expression \((X - \mu)^2\) is the definition of variance, which is \(\sigma^2\). Therefore, the expectation is simply \[1\].

Compute Expectation with the Exponential

Lastly, we compute \[ E\left\{\exp \left[t\left(\frac{X-\mu}{\sigma}\right)\right]\right\}\]. We see that this is equivalent to \(E(e^{t Z}), -h \sigma < t < h \sigma\) where \(Z = \frac{X - \mu}{\sigma}\). Using the definition of a moment generating function and referring back to Z, we can substitute Z in place of X in the definition of the moment generating function, \(M(t)\), of X, while also multiplying by the exponential of \(-t\mu/\sigma\) to account for the difference in mean, simplifying to \(e^{-\mu t / \sigma} M \left(\frac{t}{\sigma}\right)\).