Question

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

The moment generating function (mgf) of X is \( M(t) = e^{\frac{1}{2}t^2} \) and the probability density function (pdf) of X is \( f(x) = \frac{1}{\sqrt{2π}} e^{-\frac{x^2}{2}} \).

Step-by-step solution

Identify the Properties of a Normal Distribution

The moment generating function (mgf) of a Normal distribution with mean μ and variance σ^2 is given by \( M(t) = e^{\mu t + \frac{1}{2} t^{2} σ^{2}} \). The term \( E\left(X^{2 m}\right)=\frac{(2 m) !}{2^{m} m !}\) for \( m = 1, 2, 3, \ldots \) has a form of the moments of a standard normal distribution (μ=0, σ=1), because the moments of a standard normal distribution is either 0 (for odd moments) or the double factorial (for even moments).

Derive the Moment Generating Function (mgf)

The moment generating function (mgf) of a random variable X is given by: \( M(t) = E(e^(tX)) \). For a standard normal distribution (μ=0, σ=1), the expected value E(X)=μ=0 and the variance Var(X)=σ^2=1. Therefore, the mgf for a standard normal distribution X~N(0,1) becomes \( M(t) = e^{\frac{1}{2}t^2} \).

Find the Probability Density Function (pdf)

The pdf of a standard normal distribution is given by \( f(x) = \frac{1}{\sqrt{2π}} e^{-\frac{x^2}{2}} \). This is derived from the formula of pdf of a general normal distribution by substituting μ=0 and σ=1. Hence, the pdf for X which follows standard normal distribution becomes the same with these substitutions.

Key concepts

Moment Generating Function

The moment generating function (mgf) is a powerful tool in probability theory that encapsulates all the moments of a random variable into one function. To grasp this, imagine the mgf as a multi-purpose utility knife that, with just one expression, can slice out any desired moment when the proper technique is applied. Specifically, the mgf of a random variable, represented as M(t), is the expected value of the exponential function raised to the power of the product of the variable and a parameter t, mathematically given as \( M(t) = E(e^{tX}) \).

For instance, in the exercise, we are working with a random variable X that is normally distributed. The mgf of a standard normal distribution, which has a mean (μ) of 0 and a variance (σ²) of 1, simplifies to \( M(t) = e^{\frac{1}{2}t^2} \). This simplified expression reveals the underlying symmetry of the standard normal distribution - even moments are non-zero and odd moments vanish to zero. By utilizing the properties of the mgf, one can, in theory, calculate any moment of the random variable, a testament to its utility in probability and statistics.

Probability Density Function

The probability density function (pdf) serves as the backbone for understanding continuous random variables. Where the moment generating function captures moments, the pdf lays out the likelihood of the random variable assuming any given value within a range. The pdf for a standard normal distribution, which reflects the bell-curve shape commonly associated with Gaussian distributions, is articulated as \( f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \).

Peeling back the layers of this equation, the \(\frac{1}{\sqrt{2\pi}}\) is a normalizing factor ensuring the total area under the curve equals 1, a basic probability requirement. The exponential component, \( e^{-\frac{x^2}{2}} \), indicates how the probabilities diminish as we move away from the mean (in this case, zero). This density function is integral for calculating probabilities, such as the chance that the variable falls within a certain interval.

Random Variable Moments

Moments of a random variable are akin to 'snapshots' of its essence, describing its various tendencies and characteristics. Just as a series of photographs can paint a portrait of an individual's life, mathematical moments tell the story of a random variable's distribution. There are several moments that statisticians commonly consider, such as the first moment (mean) which depicts the location, the second moment (variance) which portrays the spread, and higher moments like skewness (third moment) and kurtosis (fourth moment) that illustrate the shape's tailedness and peakedness, respectively.

In the context of our exercise, the moments of a standard normal distribution devoid of odd powers (\(E(X^{2m-1}) = 0\)) and present in even powers, match the moments of the standard normal distribution, confirming its symmetrical property. This structure simplifies the explanation behind why certain moments are zero (symmetry) and others follow a specific pattern (the moments formula given in the question), ultimately demystifying the distribution's behavior through its moments.