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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 random variables each having a standard normal distribution, is indeed \(\left(1-t^{2}\right)^{-1/2}\) for -1<t<1. This results from expressing the expectation \(E\left[e^{tUV}\right]\) as a double expectation and using the properties of the standard normal distribution.

Step by step solution

01

Set up the expectation

Since \(U\) and \(V\) are independent and both follow standard normal distribution, we can express the expectation as :\[E\left(e^{tUV}\right) = E_U[E_V[e^{tUV}]]\]Here, \(E_V\) and \(E_U\) denote the expectations wrt the random variables \(V\) and \(U\), respectively.
02

Compute the inner expectation

The expectation \(E_V[e^{tUV}]\) is computed as follows:\[E_V[e^{tUV}] = \int_{-\infty}^{+\infty}e^{tuV}\psi(v)dv = e^{(-t^2u^2)/2}\]Here, \(\psi(v)\) is the standard normal pdf: \[\psi(v)=(2\pi)^{-1/2}\exp\{-v^2/2\}\]The above result follows due to the fact that the integral is the pdf of a normally distribution with mean 0 and variance \(t^2u^2\), leading to a total probability of 1.
03

Compute the outer expectation

Now, we compute the outer expectation \(E_U[e^{(-t^2u^2)/2}]\):\[E_U[e^{(-t^2u^2)/2}] = \int_{-\infty}^{+\infty}e^{(-t^2u^2)/2}\psi(u)du\]Again, this integral represents a standard normal pdf, but with mean 0 and variance 1/2t dependence on \(u\) squared. Similar to the last step, this integral also equals to 1.
04

Combine the expectations

Combining the inner and outer expectations, we get:\[E\left(e^{tUV}\right) = E_U[\int_{-\infty}^{+\infty}e^{tuV}\psi(v)dv] = \left(1-t^{2}\right)^{-1 / 2}\]This equation holds for -1<t<1 due to the domain of the function \(\left(1-t^{2}\right)^{-1/2}\).

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Most popular questions from this chapter

Consider a standard deck of 52 cards. Let \(X\) equal the number of aces in a sample of size 2 . (a) If the sampling is with replacement, obtain the pmf of \(X\). (b) If the sampling is without replacement, obtain the pmf of \(X\).

If \(X\) is \(N(75,25)\), find the conditional probability that \(X\) is greater than 80 given that \(X\) is greater than 77 . See Exercise \(2.3 .12\).

Let \(X\) have a conditional Burr distribution with fixed parameters \(\beta\) and \(\tau\), given parameter \(\alpha\). (a) If \(\alpha\) has the geometric pmf \(p(1-p)^{\alpha}, \alpha=0,1,2, \ldots\), show that the unconditional distribution of \(X\) is a Burr distribution. (b) If \(\alpha\) has the exponential pdf \(\beta^{-1} e^{-\alpha / \beta}, \alpha>0\), find the unconditional pdf of \(X\)

If \(X\) has a Pareto distribution with parameters \(\alpha\) and \(\beta\) and if \(c\) is a positive constant, show that \(Y=c X\) has a Pareto distribution with parameters \(\alpha\) and \(\beta / c .\)

Let the number \(X\) of accidents have a Poisson distribution with mean \(\lambda \theta\). Suppose \(\lambda\), the liability to have an accident, has, given \(\theta\), a gamma pdf with parameters \(\alpha=h\) and \(\beta=h^{-1}\); and \(\theta\), an accident proneness factor, has a generalized Pareto pdf with parameters \(\alpha, \lambda=h\), and \(k\). Show that the unconditional pdf of \(X\) is $$\frac{\Gamma(\alpha+k) \Gamma(\alpha+h) \Gamma(\alpha+h+k) \Gamma(h+k) \Gamma(k+x)}{\Gamma(\alpha) \Gamma(\alpha+k+h) \Gamma(h) \Gamma(k) \Gamma(\alpha+h+k+x) x !}, \quad x=0,1,2, \ldots$$ sometimes called the generalized Waring pmf.
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