# Exponential parameter estimation for exponentially modified gaussian

According to this Wikipedia article:

https://en.wikipedia.org/wiki/Exponentially_modified_Gaussian_distribution

One way to estimate the exp_gaussian parameteres given a data sample is with the method of moments. For example, to estimate the exponential decay term, $$τ$$ :

$$τ = s \cdot ( \frac{γ_{1}}{2} ) ^ {1/3}$$

Where $$γ_{1}$$ is the skewness and $$s$$ the standard deviation.

If we take a random sample (too small for estimation, but still) $$[957,1,5,271,12,85, 12,658,4719]$$:

$$m = 746.66$$

$$s = 1528.118$$

$$γ_{1} = 2.254$$

$$τ = 1590.351$$

I would like to know several things. Are this results correct (particularly for the exponential term)? What does the τ value mean graphically (what happens when it's negative, positive, etc)?

• You've given a wrong link. You want en.wikipedia.org/wiki/…. Your calculations appear correct. To find out what happens when you vary $\tau$, try plotting the pdf with varying values of $\tau$.
– JimB
Commented Oct 13, 2021 at 16:55
• Yes, I hadn't noticed. Thanks Commented Oct 14, 2021 at 14:40

The probability density function of an exponentially modified gaussian is given by this equation:

$$f(x;μ,σ,λ) = \frac{λ}{2}e^{\frac{λ}{2}(2μ+λσ^2-2x)} erfc(\frac{μ+λσ^2-x}{\sqrt{2}σ})$$

Where erfc is the complementary error function or:

$$erfc(x) = 1 - erf(x)$$

The python library SciPy has implemented erf as scipy.special.erf(). We can use it with Matplotlib to represent how the function changes by varying λ (which has the same value as $$\frac{1}{τ}$$):

EMG for different λ

My interpretation of this is that, as τ increases, a higher percentage of the distribution is accumulated in the right end of the distribution. τ is exponent relaxation time, meaning a lower τ would make the distribution reach it's horizontal asymptote much faster than a higher τ.

To confirm this practically, I created 2 lists that contained 20000 groups of numbers. In the first list I added the group [-2,2] while in the second list I added the group [4,-1,-1,-1,-1]. After calculating the parameters of this 2 lists, I get:

$$m_1 = m_2 = 0$$

$$s_1 = s_2 = 2$$

$$τ_1 = 0.027$$

$$τ_2 = 4229.471$$

Given that $$m_2 = 0$$ and $$s_2 = 2$$, 4 is 2 standard deviations above the mean. By adding [4,-1,-1,-1,-1] n-times we have that 20% of the data is in the far right of the distribution (in a normal distribution only ≈ 2% would be there). Because of this, τ is higher (feel free to comment).