Chapter 1: Problem 24
Let \(X\) be a random variable with a pdf \(f(x)\) and mgf \(M(t)\). Suppose \(f\) is symmetric about \(0,(f(-x)=f(x))\). Show that \(M(-t)=M(t)\).
Short Answer
Expert verified
We have proven that for a random variable \(X\) with a symmetric probability density function \(f(x)\), its moment generating function satisfies \(M(-t) = M(t)\).
Step by step solution
01
Definition of Moment Generating Function (MGF)
We start with the general definition of the moment generating function \(M(t)\) of a random variable \(X\) with a probability density function \(f\), which is given by: \[ M(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx}f(x)dx\].
02
Compute \(M(-t)\) in terms of \(f(x)\)
Next, we calculate \(M(-t)\) similarly, it is given by: \[ M(-t) = E[e^{-tX}] = \int_{-\infty}^{\infty} e^{-tx}f(x)dx\]. Now, to utilize the given symmetry property of \(f\), we can make a substitution in the equation. Let \(x = -y\). Then, our integral bounds change from \(-\infty\) to \(\infty\) and the differential becomes \(dx = -dy\). Then, we have: \[M(-t) = \int_{\infty}^{-\infty} e^{ty}f(-y)(-dy)\]. Implementing the property that \(f(-y) = f(y)\) we have \[M(-t) = \int_{-\infty}^{\infty} e^{ty}f(y)dy\]
03
Equality of \(M(-t)\) and \(M(t)\)
Comparing the expressions obtained at steps 1 and 2, we can now observe that: \[ M(t) = \int_{-\infty}^{\infty} e^{tx}f(x)dx = \int_{-\infty}^{\infty} e^{ty}f(y)dy = M(-t)\], thus proving that \(M(-t) = M(t)\) for a given random variable \(X\) whose probability density function \(f\) is symmetric about zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Density Function
In probability theory, a Probability Density Function (PDF) describes the likelihood of a continuous random variable taking on a particular value. Unlike probabilities in a discrete distribution, where specific values have associated probabilities, a PDF gives probability over an interval. The area under the PDF curve between two points gives the probability that the variable is within that range.
A key characteristic of PDFs is that they are non-negative and integrate to one over the entire space. This means:
A key characteristic of PDFs is that they are non-negative and integrate to one over the entire space. This means:
- For any continuous random variable, the integral of its PDF over all possible values is 1.
- The value of the PDF at any specific point can be greater than 1, but the area under the curve must always sum to 1.
Symmetric Function
A symmetric function in the context of probability refers to a function that remains unchanged even when its input sign is reversed. In mathematical terms, a function \(f(x)\) is symmetric about zero if \(f(-x) = f(x)\).
When a probability density function (PDF) is symmetric, it indicates that the likelihood of a random variable taking on positive or negative values of the same magnitude is identical.
When a probability density function (PDF) is symmetric, it indicates that the likelihood of a random variable taking on positive or negative values of the same magnitude is identical.
- This is common in distributions like the normal distribution, which is symmetric around its mean.
- Symmetry simplifies mathematical evaluations such as integrals, as seen in moment generating functions (MGF).
Random Variable
A random variable is a variable that takes on values determined by the outcome of a random phenomenon. It bridges the gap between a real-world random process and a mathematical model.
- Random variables can be discrete or continuous.
- In this context, we're focusing on continuous random variables, which have associated probability density functions (PDFs) to describe their distributions.
Expected Value
The expected value of a random variable is the long-term average or mean value it takes on over many trials. It embodies the concept of mathematical expectation, giving us a central tendency of the variable.
For a continuous random variable with a probability density function (PDF) \(f(x)\), the expected value \(E[X]\) is calculated as:\[ E[X] = \int_{-\infty}^{\infty} x f(x) \, dx \]The expected value is crucial for calculating the moment generating function (MGF), which is based on expected values of exponential transformations of the random variable.
For a continuous random variable with a probability density function (PDF) \(f(x)\), the expected value \(E[X]\) is calculated as:\[ E[X] = \int_{-\infty}^{\infty} x f(x) \, dx \]The expected value is crucial for calculating the moment generating function (MGF), which is based on expected values of exponential transformations of the random variable.
- The MGF is \(M(t) = E[e^{tX}]\), where \(t\) is any real number.
- MGFs are utilized to find all moments of a distribution, such as the mean (first moment) and variance (second moment).