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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?

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    $\begingroup$ 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
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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:

library(e1071)
library(metRology)   
df <- 8
set.seed(111) 

#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
c(mean(b),sd(b))

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

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

The result is:

c(mean(b),sd(b))
## [1] 1.4602974 0.1380863
E41
## [1] 4.5
E40
## [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.

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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:

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

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