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Below is an equation $f(x)$ that is the sum of two lognormal waveforms. All of the coefficients are known, and by integration, the area under the curve is simply $P_1 + P_2$.

$\displaystyle f(x) = \frac {P_1 } {S_1(x + D) \sqrt { 2 \Pi}} e ^ {-(\ln(x + D)-M_1) ^2 / 2S_1^2} + \frac {P_2 } {S_2(x + D) \sqrt { 2 \Pi}} e ^ {-(\ln(x + D)-M_2) ^2 / 2S_2^2} + B$

Treating these two waveforms as probability distributions, the skewness and kurtosis of the individual waveforms can be calculated from the coefficients $S_1$ and $S_2$ as

Skewness = $(e^{S_x^2}\!\!+2) \sqrt{e^{S_x^2}\!\!-1}$

Kurtosis = $\displaystyle e ^ {4S_x^2} + 2e ^ {3S_x^2} + 3e ^ {2S_x^2} - 6$

If $f(x)$ itself represents a probability distribution, how can one obtain the skewness and kurtosis of $f(x)$ from these same coefficients?

Graph of the above, showing the individual lognormal waveforms

zax
  • 81

1 Answers1

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Long story short: I worked out the solution.

Disregarding the measurement baseline B and shared delay D above, we have the following pdf: $$ f_X(x) = \frac {1} {P_1 + P_2} \left ( \frac {P_1 } {S_1x \sqrt { 2 \pi}} e ^ {-(\ln(x)-M_1) ^2 / (2S_1^2)} + \frac {P_2 } {S_2x \sqrt { 2 \pi}} e ^ {-(\ln(x)-M_2) ^2 / (2S_2^2)} \right ) $$

Using the same substitution method given here for the single lognormal pdf, the mgf of the dual lognormal pdf can be calculated as:

$$ {\rm E}[X^k] = \frac {P_1} {P_1 + P_2} e^{k(2M_1 + kS_1^2)/2} + \frac {P_2} {P_1 + P_2} e^{k(2M_2 + kS_2^2)/2} $$

Once, the mgf is known, the moments can be trivially calculated, noting $ \mu_2 = \sigma^2 $, γ1 = $ \mu_3 / \sigma^3 $ and κ1, is equal to $ \mu_4 / \sigma^4 $.

zax
  • 81