Question

Let \(U\) and \(V\) be independent random variables, each having a standard normal distribution. Show that the mgf \(E\left(e^{t(U V)}\right)\) of the random variable \(U V\) is \(\left(1-t^{2}\right)^{-1 / 2},-1

Short answer

The moment generating function (mgf) of the random variable \(UV\), where \(U\) and \(V\) are independent normal distributions, is \((1 - t^{2})^{-1/2}\) for \(-1< t< 1\), when the integral of a bivariate normal pdf with zero means is compared with \(E\left(e^{tUV}\right)\).

Step-by-step solution

Expression for Expected Value

First, compute the expected value of \(e^{tUV}\). We have \(E\left(e^{tUV}\right)=E_{U}\left[E_{V}\left(e^{tUV}\right)\right]\) due to the independence of \(U\) and \(V\). When \(t = 0\), this simplifies to \(E\left(e^{tUV}\right)=E_{U}\left[E_{V}\left(1\right)\right]=1\). For \(t ≠ 0\), we have \(E\left(e^{tUV}\right)=E_{U}\left[\int^{∞}_{-∞}e^{tUv}ϕ(v)dv\right]\), where \(ϕ(v)\) is the standard normal pdf..

Dealing with the Integral

The integral is a Gaussian Integral so it simplifies to: \(∫^{∞}_{-∞} e^{tUv}ϕ(v) dv = e^{t^{2}U^{2}/2}\) where \(ϕ(v) = 1/√{2π}e^{-v^{2}/2}\) is the standard normal pdf. So the expected value simplifies to: \(E\left(e^{tUV}\right) = E\left[e^{t^{2}U^{2}/2}\right]\) for \(t ≠ 0\).

Apply the Results to the Variables

Substitute \(Z = tU\), then, since \(U\) is Normal(0,1), \(Z\) is also Normal(0,\(t^{2}\)). Thus, \(E\left(e^{tUV}\right) = E\left[e^{Z^{2}/2}\right]\) for \(t ≠ 0 \). Now use the result \(E\left[e^{kZ^{2}/2}, -∞<k<1\right] = (1 - k)^{-1/2}\) and set \(k = t^{2}\), we get \(E\left[e^{tUV}\right] = (1 - t^{2})^{-1/2}\) for \(-1< t< 1\) and \(E\left(e^{tUV}\right) = 1\) for \(t = 0\). Therefore, the moment generating function of \(UV\) is \((1 - t^{2})^{-1/2}\) for \(-1< t< 1\).