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To address the problem, we need to demonstrate that if the probability density function (pdf) fX(x) of a continuous random variable X is descending, then its median m is greater than or equal to its mean μ.
Concavity of the CDF: The CDF FX(x) is concave if fX(x) is descending. This is because the second derivative FX′′(x)=−fX′(x)>0.
Jensen's Inequality: For a concave function g, g(E[X])≤E[g(X)]. Applying this:
FX(E[X])≤E[FX(X)]
Since FX(X) is uniformly distributed between 0 and 1, E[FX(X)]=0.5.
Interpretation: Since FX(E[X])≤0.5, the median m, which is defined by FX(m)=0.5, must be greater than or equal to the mean μ.
The descending nature of fX(x) implies a right-skewed distribution, where the mean μ 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≥μ holds true.