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Let \(X\) be a random variable such that \(E\left(X^{2 m}\right)=(2 m) ! /\left(2^{m} m !\right), m=\) \(1,2,3, \ldots\) and \(E\left(X^{2 m-1}\right)=0, m=1,2,3, \ldots\) Find the mgf and the pdf of \(X\).

Short Answer

Expert verified
The moment generating function (mgf) of \(X\) is \(M(t) = e^{t^2}\), and the probability density function (pdf) is \(f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\). This indicates that the random variable \(X\) follows a standard normal distribution.

Step by step solution

01

Establishing Variable Representations

For random variable \(X\), given that \(E(X^{2m}) = \frac{(2m)!}{2^m m!}\) and \(E(X^{2m-1}) = 0\), where \(m\) belongs to the natural numbers. It's important to recognize that the expectations for even powers of \(X\) are given by the ratio of two factorials, whereas for odd powers the expected value equals zero.
02

Constructing the Moment Generating Function (MGF)

By definition, the moment generating function, \(M(t)\), of a random variable \(X\) is \( M(t) = E(e^{tX})\) . The mgf is basically a power series, expressed as \(M(t) = 1 + tE(X) + \frac{t^2E(X^2)}{2!} + \frac{t^3E(X^3)}{3!} +...\) . Since all odd powered expectations equal zero, the odd powered terms of the power series will vanish and only the even terms will remain. So, the moment generating function after putting in the known values would be \(M(t) = 1 + 0 + \frac{t^2(2!) / 2^1 1! }{2!} + 0 + \frac{t^4(4!) / 2^2 2! }{4!} +... = 1 + t^2 + \frac{t^4}{2^2} +...\), which is the series representation of \(e^{t^2}\). So, \(M(t) = e^{t^2}\).
03

Determine the Probability Density Function (PDF)

A well-known fact in probability theory is that the distribution of a random variable is uniquely characterized by its moment generating function. Hence, recognizing that the mgf \(M(t)=e^{t^2}\) is that of a standard normal distribution, it can be concluded that \(X\) follows a standard normal distribution. So, the pdf of \(X\) is given by \(f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\).

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