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Chapter 3: Problem 4

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

Expert verified
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

01

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 \(t 鈮� 0\), we have \(E\left(e^{tUV}\right)=E_{U}\left[\int^{鈭瀩_{-鈭瀩e^{tUv}蠒(v)dv\right]\), where \(蠒(v)\) is the standard normal pdf..
02

Dealing with the Integral

The integral is a Gaussian Integral so it simplifies to: \(鈭玘{鈭瀩_{-鈭瀩 e^{tUv}蠒(v) dv = e^{t^{2}U^{2}/2}\) where \(蠒(v) = 1/鈭歿2蟺}e^{-v^{2}/2}\) 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 \(t 鈮� 0\).
03

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 \(t 鈮� 0 \). Now use the result \(E\left[e^{kZ^{2}/2}, -鈭�<k<1\right] = (1 - k)^{-1/2}\) 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 \(-1< t< 1\).

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Most popular questions from this chapter

. In the Poisson postulates of Remark 3.2.1, let \(\lambda\) be a nonnegative function of \(w\), say \(\lambda(w)\), such that \(D_{w}[g(0, w)]=-\lambda(w) g(0, w) .\) Suppose that \(\lambda(w)=\) \(k r w^{r-1}, r \geq 1\) (a) Find \(g(0, w)\) noting that \(g(0,0)=1\). (b) Let \(W\) be the time that is needed to obtain exactly one change. Find the distribution function of \(W\), i.e., \(G(w)=P(W \leq w)=1-P(W>w)=\) \(1-g(0, w), 0 \leq w\), and then find the pdf of \(W\). This pdf is that of the Weibull distribution, which is used in the study of breaking strengths of materials.

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