Question

Let \(p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere, be the pmf of the random variable \(X\). Find the mgf, the mean, and the variance of \(X\).

Short answer

The moment generating function (mgf) of \(X\) is \(M(t) = \frac{e^t}{2-e^t}\), the mean of \(X\) is \(0.5\), and the variance of \(X\) is \(0.75\).

Step-by-step solution

Moment Generating Function Calculation

The moment generating function \(M(t)\) is defined by the expected value of \(e^{tX}\). For a discrete random variable, \(M(t)\) can be calculated as:\n\n\(M(t) = \sum_{x=1}^{\infty}e^{tx}p(x)\)\n\nSubstituting \(p(x) = \left(\frac{1}{2}\right)^{x}\) into the equitation gives us:\n\n\(M(t) = \sum_{x=1}^{\infty}e^{tx}\left(\frac{1}{2}\right)^{x} = \sum_{x=1}^{\infty}\left(\frac{e^t}{2}\right)^{x}.\)\n\nThis is a geometric series with common ratio \(r=\frac{e^t}{2}\), so it converges when \(|r| < 1\), i.e., when \(-\ln(2)<t<\ln(2)\). The sum of the series is then given by:\n\n\(M(t) = \frac{\frac{e^t}{2}}{1-\frac{e^t}{2}} = \frac{e^t}{2-e^t}\).

Calculation of the Mean

The mean of a random variable is the first moment about the origin, \(E[X]\), and can be obtained by differentiating the moment generating function with respect to \(t\) and evaluating at \(t=0\), because \(M'(t)|_{t=0} = E[X]\). So, we differentiate \(M(t) = \frac{e^t}{2-e^t}\) to get:\n\n\(M'(t) = \frac{2e^{2t}}{(2-e^t)^2}\).\n\nEvaluating at \(t=0\), we get:\n\n\(M'(0) = E[X] = \frac{2}{4} = \frac{1}{2}\).

Calculation of the Variance

The variance of a random variable is the second central moment, \(E[X^2] - (E[X])^2\), and can be obtained by differentiating the moment generating function twice with respect to \(t\) and evaluating at \(t=0\), because \(M''(t)|_{t=0} = E[X^2]\). Differentiating \(M'(t) = \frac{2e^{2t}}{(2-e^t)^2}\) we get:\n\n\(M''(t) = \frac{8e^{3t}}{(2-e^t)^3}\).\n\nEvaluating at \(t=0\), we get \(M''(0) = E[X^2] = 2\). Therefore, the variance is given by:\n\nVar(X) = E[X^2] - (E[X])^2 = 2 - \left(\frac{1}{2}\right)^2 = 2 - \frac{1}{4} = \frac{3}{4}\).\n