Common parameters for probability distribution include the mean and standard deviation. The mean gives a measurement of the center and the standard deviation tells how spread out the distribution is. In addition to these well-known parameters, there are others that draw attention to features other than the spread or the center. One such measurement is that of skewness. Skewness gives a way to attach a numerical value to the asymmetry of a distribution.

One important distribution that we will examine is the exponential distribution. We will see how to prove that the skewness of an exponential distribution is 2.

### Exponential Probability Density Function

We begin by stating the probability density function for an exponential distribution. These distributions each have a parameter, which is related to the parameter from the related Poisson process. We denote this distribution as Exp(A), where A is the parameter. The probability density function for this distribution is:

*f*(*x*) = *e*^{-x/A}/A, where *x* is nonnegative.

Here *e* is the mathematical constant *e* that is approximately 2.718281828. The mean and standard deviation of the exponential distribution Exp(A) are both related to the parameter A. In fact, the mean and standard deviation are both equal to A.

### Definition of Skewness

Skewness is defined by an expression related to the third moment about the mean.

This expression is the expected value:

E[(X – μ)^{3}/σ^{3}] = (E[X^{3}] – 3μ E[X^{2}] + 3μ^{2}E[X] – μ^{3})/σ^{3} = (E[X^{3}] – 3μ(σ^{2} – μ^{3})/σ^{3}.

We replace μ and σ with A, and the result is that the skewness is E[X^{3}] / A^{3} – 4.

All that remains is to calculate the third moment about the origin. For this we need to integrate the following:

∫^{∞}_{0} *x* ^{3} *f*(*x*) d*x*.

This integral has an infinity for one of its limits. Thus it can be evaluated as a type I improper integral. We also must determine what integration technique to use. Since the function to integrate is the product of a polynomial and exponential function, we would need to use integration by parts. This integration technique is applied several times. The end result is that:

E[X^{3}] = 6A^{3}

We then combine this with our previous equation for the skewness. We see that the skewness is 6 – 4 = 2.

### Implications

It is important to note that the result is independent of the specific exponential distribution that we start with. The skewness of the exponential distribution does not rely upon the value of the parameter A.

Furthermore, we see that the result is a positive skewness. This means that the distribution is skewed to the right. This should come as no surprise as we think about the shape of the graph of the probability density function. All such distributions have y-intercept as 1//theta and a tail that goes to the far right of the graph, corresponding to high values of the variable *x*.

### Alternate Calculation

Of course, we should also mention that there is another way to calculate skewness.

We can utilize the moment generating function for the exponential distribution. The first derivative of the moment generating function evaluated at 0 gives us E[X]. Similarly, the third derivative of the moment generating function when evaluated at 0 gives us E(X^{3}].