Question

Show that the moment generating function of the random variable \(X\) having the pdf \(f(x)=\frac{1}{3},-1

Short answer

The moment generating function is confirmed: \( M(t) = \frac{e^{2t}-e^{-t}}{3t} \) for \( t \neq 0 \) and \( M(t) = 1 \) for \( t = 0 \).

Step-by-step solution

Definition of MGF

By definition, the moment generating function (MGF) of a random variable \(X\) is given by \(M(t) = E[e^{tX}]\) where \(E\) is the expectation of the random variable.

The Expected Value Calculation

To find \(E[e^{tX}]\), integrate the product of the pdf and the exponential function across the range of \(X\), which is \( -1 < X < 2\). This will give: \[E[e^{tX}]= \int_{-1}^{2} e^{tx}f(x) dx = \int_{-1}^{2} e^{tx}\frac{1}{3} dx \] Since \( f(x) = \frac{1}{3} \) for \( -1 < X < 2 \) and 0 elsewhere.

Calculation of the Integral

Calculate the integral from Step 2: \[=\frac{1}{3} \int_{-1}^{2} e^{tx} dx = \frac{e^{tx}}{3t} \Big |_{-1}^{2} = \frac{e^{2t}-e^{-t}}{3t} \] However, the above solution is for \(t \neq 0\). For \( t=0 \), the integral seems undefined, but in fact, by using l'Hôpital's rule, when \(t=0\), it can be shown that the right limit results a finite number: \( \lim_{t\to 0}\frac{e^{2t}-e^{-t}}{3t} = 1. \) So, we can define the MGF as \( M(t) = \frac{e^{2t}-e^{-t}}{3t} \) for \( t \neq 0 \) and \( M(t) = 1 \) for \( t = 0 \). Hence, this proves the proposed MGF.