Question

Let \(\psi(t)=\log M(t)\), where \(M(t)\) is the mgf of a distribution. Prove that \(\psi^{\prime}(0)=\mu\) and \(\psi^{\prime \prime}(0)=\sigma^{2} .\) The function \(\psi(t)\) is called the cumulant generating function.

Short answer

The first derivative of the cumulant generating function at t=0 equals the mean (\(\mu\)) of the distribution, and the second derivative at t=0 equals the variance (\(\sigma^2\)).

Step-by-step solution

Understand the Requirement

We need to prove that the first derivative of the cumulant generating function, \(\psi(t) = \log{M(t)}\), evaluated at t=0, equals the mean (\(\mu\)) of a distribution, and the second derivative evaluated at t=0 equals the variance (\(\sigma^{2}\)).

Compute the First Derivative

We compute the first derivative of \(\psi(t)\), which we will denote as \(\psi^{\prime}(t)\). By using the chain rule, we obtain \(\psi^{\prime}(t) = \frac{M^{\prime}(t)}{M(t)}\). Then, we evaluate this at t=0, i.e. \(\psi^{\prime}(0) = \frac{M^{\prime}(0)}{M(0)}\).

Use Properties of Moment Generating Function

The moment generating function evaluated at t=0, i.e. M(0), is always 1 for any distribution. This gives us \(\psi^{\prime}(0) = M^{\prime}(0)\). It is also known that the first derivative of the moment generating function evaluated at t=0 is equal to the mean, \(\mu\), of the distribution. Hence \(\psi^{\prime}(0) = M^{\prime}(0) = \mu\).

Compute the Second Derivative

Next, we need to compute the second derivative of \(\psi(t)\), denoted as \(\psi^{\prime\prime}(t)\). We differentiate \(\psi^{\prime}(t)\) to get \(\psi^{\prime\prime}(t) = \frac{M^{\prime\prime}(t)}{M(t)} - \left(\frac{M^{\prime}(t)}{M(t)}\right)^2\). Then, we evaluate this at t=0, i.e. \(\psi^{\prime\prime}(0) = \frac{M^{\prime\prime}(0)}{M(0)} - \left(\frac{M^{\prime}(0)}{M(0)}\right)^2\).

Use Properties again of Moment Generating Function

Again, since M(0) is 1, we get \(\psi^{\prime\prime}(0) = M^{\prime\prime}(0) - \left(M^{\prime}(0)\right)^2\). We know that \(\psi^{\prime}(0) = M^{\prime}(0) = \mu\), and \(M^{\prime\prime}(0) = \mu^2 + \sigma^2\). Substituting these values into the formula, we have \(\psi^{\prime\prime}(0) = \mu^2 + \sigma^2 - \mu^2\), which simplifies to \(\psi^{\prime\prime}(0) = \sigma^2\).