Question

Suppose \(X\) is a random variable with the pdf \(f(x)\) which is symmetric about \(0 ;\) i.e., \(f(-x)=f(x) .\) Show that \(F(-x)=1-F(x)\), for all \(x\) in the support of \(X\).

Short answer

Proven that for a random variable X with a symmetric pdf, \(F(-x)=1-F(x)\) holds true.

Step-by-step solution

- Understand cumulative distribution function (CDF)

The CDF, \(F(x)\), of a random variable \(X\) is defined as the area under its pdf from \(-\infty\) to \(x\). Mathematically, \(F(x) = \int_{-\infty}^{x} f(t) dt\). We need to utilize this definition to prove the given expression.

- Express \(F(-x)\) using definition of CDF

Applying the definition of the CDF to the given problem, we can write the CDF of \(-x\) as \(F(-x) = \int_{-\infty}^{-x} f(t) dt\) since the pdf \(f(x)\) is symmetric.

- Apply the symmetry of f(x) = f(-x)

Since the function is symmetric about the origin, we can replace \(t\) in the integral with \(-t\). This gives us \(F(-x) = \int_{-\infty}^{-x} f(-t) dt\)

- Change of variables and limits

Make a change of variable by letting \(t = -u\). This results in \(dt = -du\) and the limits of the integral will change to positive infinity to x. By substitifying these into the integral, we get \(F(-x) = \int_{\infty}^{x} -f(u) du = -\int_{x}^{\infty} f(u) du\)

- Show that \(F(-x)=1-F(x)\)

Remembering that \( \int_{-\infty}^{\infty} f(x) dx = 1 \), because the total probability should add up to 1, we can express \(1 = \int_{-\infty}^{x} f(u) du + \int_{x}^{\infty} f(u) du\). Rearranging for \(-\int_{x}^{\infty} f(u) du\) gives us \(1 - \int_{-\infty}^{x} f(u) du = -\int_{x}^{\infty} f(u) du\). By comparing with the result from Step 4, we find that \(F(-x)= 1 - F(x)\)

Key concepts

Probability Density Function

The Probability Density Function (PDF) is a statistical expression defining a probability distribution for a continuous random variable. Unlike a cumulative distribution function (CDF), the PDF describes the likelihood for this random variable to take on a particular value. More formally, for a continuous random variable X, its PDF, denoted as f(x), represents the probability that X is exactly x. However, for continuous random variables, we think in terms of probabilities of intervals rather than exact values.

The area under the curve of the PDF over a certain interval represents the probability that the random variable falls within that interval. Mathematically, to find this probability, an integral of the PDF is computed over that range of values. This is the connection between the PDF and the CDF - while the PDF gives you the 'density' of probabilities, the CDF sums up probabilities up to a certain point.

Symmetry in Probability Distributions

Symmetry in a probability distribution indicates that the distribution has mirror-image balance about a central point, known as the mean of the distribution. In a symmetric distribution, if you pick a point on one side of the mean and then find a point of equal distance on the other side, both points will have the same probability density.

This characteristic can greatly simplify analyses and solutions to probabilistic problems. For instance, in the context of the given exercise, the symmetry of a probability distribution about zero implies that f(-x) = f(x) for all x. Thus, probabilities calculated for negative values have a direct relationship to probabilities for the positive values, which we leveraged to derive that F(-x) = 1 - F(x) where F(x) is the CDF.

Integral Calculus in Statistics

Integral calculus plays a fundamental role in statistics, particularly when dealing with continuous probability distributions. While the PDF provides the density function of a random variable, it is the integral which accumulates these densities to provide probabilities and expected values.

For continuous distributions, the CDF is obtained by integrating the PDF. This is because integration accumulates the 'density' of probabilities across an interval, answering the question of 'how likely is it that the variable falls below a certain value?'. Moreover, important statistical measures such as the mean and variance can also be derived via integrals of the PDF. In the exercise we've discussed, integral calculus is used to transform the PDF into the CDF and to showcase the relationship between the probabilities of opposite points in a symmetric distribution.