I am considering the unit-variance t-distribution.

I have read that the fourth moment in such case is given by 3(v-2)/(v-4) where v is the degrees of freedom. Can someone explain how does this follow?

  • 3
    $\begingroup$ Well, it's math, based on calculating the expectation of $x^4$. Are you asking for the derivation? Also note that the standard t-distribution does not have unit variance, it has a scale parameter = 1, which is not the same. $\endgroup$ – jbowman Apr 20 '18 at 21:36
  • $\begingroup$ Yes, If possible $\endgroup$ – Anna Apr 20 '18 at 21:37
  • $\begingroup$ Two methods that ought to be relatively simple are (1) use the representation of the Student-t distribution as a variance mixture of normals or (2) expand its characteristic function to fourth order. You can also compute the fourth moment directly using the Calculus of Residues: there are only two poles at $\pm i/\sqrt{\nu}$ to consider. $\endgroup$ – whuber Apr 21 '18 at 13:55

As @whuber♦ commented below the question: Kurtosis of a standardized Student's-t distribution?

Kurtosis by definition is invariant under affine linear transformations, which includes standardization.

I think you may calculate the Kurtosis and times it by variance, which in your case is 1. As the Kurtosis doesn't change, the fourth moment of unit variance t distribution is still 3 + 6/(v-4), where v is the degree of freedom. This result is the same with your formula 3(v-2)/(v-4).

I validated the result by simulation in R:

df <- 8

#empty vector a to save sample variance
a <- vector(mode = "numeric", length = 50)

#empty vector b to save sample fourth moment
b <- vector(mode = "numeric", length = 50)

#kurtosis of a random variable with unit variance
 for (i in 1 : length(a)){
random.t <- rt.scaled(100000, df = 8, mean = 0, sd = sqrt((df-2)/df))

#sample variance
a[i] <- var(random.t)

#sample 4th moments
b[i] <- kurtosis(random.t)*a[i]^2
# simulated sample 4th moments - mean and sd

#the 4th moment by your formula
E41 <- 3*(df-2)/(df-4)

#"affine linear transformation doesn't change the kurtosis"
E40 <- 6/(df-4) + 3
#theoretical the 4th moments
E40 * 1^2

The result is:

## [1] 1.4602974 0.1380863
## [1] 4.5
## [1] 4.5

It is tricky that the Kurtosis formula on Wikipedia page of t distribution is the excess Kurtosis, which is Kurtosis - 3. As a result, we should add 3 to the simulated result 1.46 + 3 = 4.46, which is close to the theoretical one (4.5).

Please correct me if there is any problem.

| cite | improve this answer | |

In an answer to a related question here I show how to derive the raw moments of the T distribution using its respresentation as a mixture of normals. The result of this analysis is that the moments of the distribution exist for all orders $0<k<\varphi$ with values given by:

$$\mathbb{E}(T^k) = \begin{cases} 0 & & & \text{if } k \text{ is odd}, \\[6pt] \frac{\Gamma(\tfrac{k+1}{2})}{\sqrt{\pi}} \cdot \frac{\Gamma(\tfrac{\varphi-k}{2})}{\Gamma(\tfrac{\varphi}{2})} \cdot \varphi^{k/2} & & & \text{if } k \text{ is even}. \end{cases}$$

Application of this general formula for the even moments yields:

$$\begin{equation} \begin{aligned} \mathbb{E}(T^2) &= \frac{\varphi}{\varphi-2} & & & \text{for } \varphi > 2, \\[6pt] \mathbb{E}(T^4) &= \frac{3 \varphi^2}{(\varphi-2) (\varphi-4)} & & & \text{for } \varphi > 4. \\[6pt] \end{aligned} \end{equation}$$

Putting this together, we see that for $\varphi>4$ the kurtosis exists and is given by:

$$\begin{equation} \begin{aligned} \mathbb{Kurt}(T) = \frac{\mathbb{E}(T^4)}{\mathbb{E}(T^2)^2} &= \frac{3 \varphi^2}{(\varphi-2) (\varphi-4)} \Big/ \frac{\varphi^2}{(\varphi-2)^2} \\[6pt] &= \frac{3}{\varphi-4} \Big/ \frac{1}{\varphi-2} \\[6pt] &= 3 \cdot \frac{\varphi-2}{\varphi-4} . \\[6pt] \end{aligned} \end{equation}$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.