Question

We say that \(X\) has a Laplace distribution if its pdf is $$ f(t)=\frac{1}{2} e^{-|t|}, \quad-\infty

Short answer

To solve the task, the integral of the pdf with the exponential function to find the mgf was calculated. This was simplified to show it matches the given mgf form, valid for \(|t|<1\). To find the moments of \(X\), the mgf was differentiated and evaluated at \(t=0\). This gives the nth moment as \(E(X^{n}) = 2n!(2n-1)!!\).

Step-by-step solution

Calculating the mgf

The mgf \(M(t)\) of a random variable \(X\) is defined as \( M(t)=E(e^{tX}) = \int_{-\infty}^{\infty} e^{tx} f(x) dx \), where \( f(x) \) is the pdf of \(X\). For the Laplace distribution, \(f(x) = \frac{1}{2} e^{-|x|}\). Splitting the integral into two parts for \( x < 0 \) and \( x >= 0 \), this becomes: \[ M(t) = \int_{-\infty}^{0} \frac{1}{2}e^{tx+x}dx + \int_{0}^{\infty} \frac{1}{2}e^{tx-x}dx \]

Simplifying the mgf

Each of the two integrals in Step 1 can be evaluated using the exponential integral rule, yielding: \[ M(t) = \frac{1}{2(1-t)} + \frac{1}{2(1+t)} = \frac{1}{1-t^{2}} \] This is valid for \( |t|<1 \), satisfying part (a) of the problem

Finding the moments using a Maclaurin series

A Maclaurin series is an approximation of a function near 0, given as \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} x^{n} \), where \( f^{(n)}(0) \) is the nth derivative of the function at 0. The nth moment of \(X\) is given as the nth derivative of \(M(t)\) at \(t=0\). Hence, each term of the Maclaurin series of \(M(t)\) gives a moment of \(X\) when evaluated at \( t=0 \). For \(M(t)=\frac{1}{1-t^{2}}\), its nth derivative is \( M^{(n)}(t) = \frac{2n!(2n-1)!!}{(1-t^{2})^{2n+1}} \), where !! denotes double factorial. Evaluating at \(t=0\), the nth moment of \(X\) becomes \(E(X^{n}) = M^{(n)}(0) = 2n!(2n-1)!!\)