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Data Interview Question

Descending Probability Density Function

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Solution & Explanation

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

Understanding the Problem

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

Explanation

1. Characteristics of a Descending pdf

  • A descending pdf implies that as xx increases, fX(x)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 FX(x)F_X(x) is concave if fX(x)f_X(x) is descending. This is because the second derivative FX(x)=fX(x)>0F_X''(x) = -f_X'(x) > 0.

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

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

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

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

4. Example: Exponential Distribution

  • Consider the exponential distribution with f(x)=λeλxf(x) = \lambda e^{-\lambda x} for x0x \geq 0.
  • The mean μ=1/λ\mu = 1/\lambda and the median m=ln(2)/λm = \ln(2)/\lambda.
  • Since ln(2)0.693\ln(2) \approx 0.693, and ln(2)<1\ln(2) < 1, the mean is greater than the median, consistent with the right-skewed nature of the distribution.

Conclusion

The descending nature of fX(x)f_X(x) implies a right-skewed distribution, where the mean μ\mu is typically greater than the median mm. 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, mμm \geq \mu holds true.