Descending Probability Density Function

Solution & Explanation

To address the problem, we need to demonstrate that if the probability density function (pdf) $f_X(x)$ of a continuous random variable $X$ is descending, then its median $m$ is greater than or equal to its mean $\mu$.

Understanding the Problem

1. Definitions:
   • Probability Density Function (pdf): A function $f_X(x)$ that describes the likelihood of $X$ taking on a particular value.
   • Mean ($\mu$): The expected value of $X$, calculated as $\int_{-\infty}^{\infty} x f_X(x) \, dx$.
   • Median ($m$): The value of $X$ for which the cumulative distribution function (CDF) $F_X(m) = 0.5$.
2. Assumption:
   • The pdf $f_X(x)$ is a descending function, meaning $f_X'(x) < 0$ for all $x$ in the support of $X$.

Explanation

1. Characteristics of a Descending pdf

• A descending pdf implies that as $x$ increases, $f_X(x)$ decreases.
• This typically results in a distribution that is right-skewed, with a longer tail on the right side.

2. Mean vs. Median in Right-Skewed Distributions

• In a right-skewed distribution, the mean is influenced by the longer tail, pulling it towards larger values.
• The median is less sensitive to the tail and remains closer to the "center" of the data.

3. Mathematical Justification

• Concavity of the CDF: The CDF $F_X(x)$ is concave if $f_X(x)$ is descending. This is because the second derivative $F_X''(x) = -f_X'(x) > 0$.

• Jensen's Inequality: For a concave function $g$, $g(E[X]) \leq E[g(X)]$. Applying this:

  $F_X(E[X]) \leq E[F_X(X)]$

  Since $F_X(X)$ is uniformly distributed between 0 and 1, $E[F_X(X)] = 0.5$.

• Interpretation: Since $F_X(E[X]) \leq 0.5$, the median $m$, which is defined by $F_X(m) = 0.5$, must be greater than or equal to the mean $\mu$.

4. Example: Exponential Distribution

• Consider the exponential distribution with $f(x) = \lambda e^{-\lambda x}$ for $x \geq 0$.
• The mean $\mu = 1/\lambda$ and the median $m = \ln(2)/\lambda$.
• Since $\ln(2) \approx 0.693$, and $\ln(2) < 1$, the mean is greater than the median, consistent with the right-skewed nature of the distribution.

Conclusion

The descending nature of $f_X(x)$ implies a right-skewed distribution, where the mean $\mu$ is typically greater than the median $m$. This is due to the influence of larger values in the tail on the mean, while the median remains more robust and central. Thus, for a strictly descending pdf, $m \geq \mu$ holds true.