Question

Let \(X\) be a random variable such that \(R(t)=E\left(e^{t(X-b)}\right)\) exists for \(t\) such that \(-h

Short answer

The mth derivative of the Moment Generating Function (MGF) of X-b at t=0 is equal to the mth moment of the distribution about the point b, which can be shown by taking the derivatives of the MGF and understanding the properties of the MGF and moments.

Step-by-step solution

Understand Moment Generating Function

We know that the Moment Generating Function (MGF) for a random variable \(X\) is represented by \(M(t) = E(e^{tX})\). For random variable \(X\) shifted by an amount \(b\), the MGF \(R(t) = E(e^{t(X-b)})\).

Derive the Moment Generating Function

We are given R(t), the MGF of (X-b). In order to find the mth moment, we need to take the mth derivative of R(t) with respect to \(t\)

Apply Principles of Derivative

The \(m\)th derivative of a MGF evaluated at \(t=0\) gives the \(m\)th moment about zero. Mathematically, that can be represented as \(R^{(m)}(0)=E[(X-b)^{m}]\).

Define Moment about the Point

Here, E[(X-b)^m] represents the \(m\)th moment of the distribution about the point \(b\). This justifies our statement given in the problem.