Question

Let \(X\) be a random variable with a pdf \(f(x)\) and mgf \(M(t)\). Suppose \(f\) is symmetric about \(0 ;\) i.e., \(f(-x)=f(x) .\) Show that \(M(-t)=M(t)\).

Short answer

The moment generating function \(M(t)\) of a random variable \(X\) with a symmetric pdf will also be symmetric: \(M(-t) = M(t)\). This is concluded from using the definition of the moment generating function and employing a variable substitution in the integral.

Step-by-step solution

Define Moment Generating Function

The moment generating function \(M(t)\) of a random variable \(X\) is defined as \(M(t) = E[e^{tX}]\). In the continuous case, this is given by the integral \(M(t) = \int_{-\infty}^{\infty} e^{tx} f(x) dx\), where \(f(x)\) is the pdf of \(X\).

Substitute the Symmetry Property of the pdf

Given that \(f\) is symmetric about 0, we know that \(f(-x) = f(x)\). We want to show that \(M(-t) = M(t)\), so we substitute this into the definition of \(M(t)\): \(M(-t) = \int_{-\infty}^{\infty} e^{-tx} f(x) dx\). Now, change the variable of integration from \(x\) to \(-x\), giving the new equation \(M(-t) = \int_{\infty}^{-\infty} e^{tu} f(u) du\), where \(u = -x\). However, as the limits of integration are now reversed, we need to change them back to \(\(-\infty, \infty\)\), which introduces a minus sign.

Final Step

This then gives \(M(-t) = \int_{-\infty}^{\infty} e^{tu} f(u) (-du) = -\int_{-\infty}^{\infty} e^{tu} f(u) du = -M(t)\). However, because the pdf \(f(x)\) is symmetric, \(f(u) = f(-u)\) and we can write \(M(-t) = \int_{-\infty}^{\infty} e^{tu} f(-u) du = M(t)\). Thus, we conclude that for a symmetric pdf, \(M(-t) = M(t)\).

Key concepts

Probability Density Function (pdf)

The Probability Density Function, commonly abbreviated as pdf, is a fundamental concept in statistics representing a random variable's distribution. It's a function that describes the likelihood of a random variable taking on a specific value or range of values. For a continuous random variable, the pdf provides the probabilities over a continuous range of possible values.

When graphed, the pdf takes the form of a curve, where the total area under the curve is equal to 1, symbolizing the fact that the probability of the random variable falling somewhere within the range is 100%. Mathematically speaking, for a random variable X with pdf denoted as f(x), the probability that X lies in an interval a to b is given by the integral of f(x) from a to b:
\[\begin{equation}\text{P}(a \< X \< b) = \int_{a}^{b} f(x) dx\end{equation}\]
Understanding the pdf is crucial for solving various types of statistical problems, such as calculating probabilities, means, and variances.

Symmetry Property in Statistics

The concept of symmetry in statistics often pertains to the shape and balance of a probability distribution. A symmetric pdf signifies that the distribution looks the same on the left and right of its central point, usually the mean. This balances the distribution such that the mean, median, and mode are the same or very close.

For example, with a symmetric pdf centered around zero, the function satisfies the condition:\[\begin{equation}f(-x) = f(x)\end{equation}\]
Mathematically, this means that for every positive value the function takes at a point x, it takes the same value at the corresponding negative point -x.

When dealing with moment generating functions (mgf), the symmetry property implies that the mgf of the negative of a symmetric random variable is the same as the original mgf, which is notably shown in our original exercise where the mgf for a symmetric pdf satisfies the equation:\(M(-t) = M(t)\).

Expected Value

The expected value, or mean, of a random variable is a measure of the central tendency or the average that a variable is expected to achieve over an infinite number of trials or samples. It essentially summarizes the distribution of possible values of the random variable by indicating where on the number line the values tend to cluster. For continuous random variables, the expected value is calculated as an integral involving the pdf.

In the context of our exercise, the expected value is used in defining the moment generating function, which in turn is denoted as \(M(t)\) for a random variable X with mean \(\mu\) and pdf f(x). It is represented mathematically as:\[\begin{equation}E[X] = \int_{-\infty}^{\infty} xf(x) dx\end{equation}\]
where E[X] stands for the expected value of X. Understanding expected value is pivotal in many fields, including economics, finance, and risk management, where it helps in making decisions under uncertainty.