Question

Let \(X\) be a random variable such that \(E\left(X^{m}\right)=(m+1) ! 2^{m}, m=1,2,3, \ldots\). Determine the mgf and the distribution of \(X\).

Short answer

The moment generating function of \(X\) is \(M_X(t) = 2t + 4t^2 + 6t^3 + \dots\) and its distribution is a scaled Exponential distribution with parameter \(\lambda = 2\).

Step-by-step solution

Identify the mgf

From the problem, the \(m\)th moment of the random variable \(X\) is given by \(E(X^m) = (m+1)!2^m. This is the \(m\)th derivative of the mgf evaluated at \(t=0\). So, by setting \(t=0\) in the Taylor expansion of the mgf \(M_X(t)\), we get: \[M_X(t) = \sum_{m=0}^{\infty} \frac{t^m E(X^m)}{m!} = \sum_{m=0}^{\infty} t^m (m + 1)! 2^m\]

Simplify the mgf

We simplify the above expression by re-indexing the sum to start from \(m=1\) instead of \(m=0\) (as the zeroth moment \[E(X^0)\] is always 1 for any random variable \(X\)) and rearranging terms:\[M_X(t) = \sum_{m=1}^{\infty} (2t)^m (m + 1) = 2t + 4t^2 + 6t^3 + \dots\]

Recognize the distribution

Now we have the form of the mgf. By looking at standard forms of mgf, we can recognize the type of the distribution. The given mgf matches with that of a scaled Exponential distribution with parameter \(\lambda = 2\). So, \(X\) follows a scaled exponential distribution.