Question

Let \(X\) be \(N(0,1)\). Use the moment generating function technique to show that \(Y=X^{2}\) is \(\chi^{2}(1)\). Hint: Evaluate the integral that represents \(E\left(e^{t X^{2}}\right)\) by writing \(w=x \sqrt{1-2 t}\), \(t<\frac{1}{2}\)

Short answer

The random variable \(Y = X^2\), where \(X\) is normally distributed \(N(0,1)\), follows a chi-square distribution with 1 degree of freedom, \(\chi^2(1)\), as shown by evaluating the moment generating function of \(Y\).

Step-by-step solution

Write out the MGF of Y

We start by writing out the MGF of \(Y\), denoted as \(M_Y(t)\). The MGF of a random variable \(Y\) can be defined as \(E\left(e^{tY}\right)\). Since \(Y = X^2\), we have \(M_Y(t) = E\left(e^{tX^2}\right)\).

Substitute \(w = x \sqrt{1-2t}\)

The next step involves substituting \(w = x \sqrt{1-2t}\) into the expectation integral. To do this, note that \(dw = \sqrt{1-2t}dx\) and \(x = \frac{w}{\sqrt{1-2t}}\). Then we substitute these values into the expectation integral, giving us \(M_Y(t) = E\left(e^{tw^2/(1-2t)}\right)*|\sqrt{1-2t}|\). This integral has the same form as the MGF of a chi-square random variable with 1 degree of freedom, except for a scaling factor \(\sqrt{1-2t}\).

Rearrange and identify the MGF

We can show that \(1-2t\) actually equals \(1^2\), and thus the scaling factor equals 1, yielding our required MGF of a chi-square random variable with 1 degree of freedom. Thus, rearranging the last equation to match the MGF of a chi-square random variable gives us \(M_Y(t) = E\left(e^{tw^2}\right)\), where \(t < \frac{1}{2}\). This implies \(Y\) indeed follows a chi-square distribution with 1 degree of freedom.