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I am looking to derive moments for several distributions. For example, I've used moment generating functions for the standard normal distribution to get:

mean =0, variance = 1, skewness = 0, kurtosis = 3

I'm looking for a similar method to do the same for the student-t distribution.

Moment generating functions are too complicated (or don't exist) for the this distribution and I've searched for an alternative on google but the only information I've found is it's extremely complicated to derive these:

mean = μ, variance = $\sigma^2 \frac{v}{v-2}$`, skewness = 0, kurtosis = 3 + $\frac{6}{v-4}$

If I have not made it clear enough than what I want to do is derive the moments shown on page 4. Both the lower moments and higher moments.

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    $\begingroup$ Mean equal to $\mu$? Are you talking about the central or non-central distribution? $\endgroup$ – JohnK Mar 24 '17 at 8:17
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    $\begingroup$ Since those are standard results, I think the self-study tag should appear in your question. First and third moments (assuming $\nu$ is large enough) are obvious by symmetry. For second and fourth moments, you can use the representation $X=\mu+\sigma\times Z/\sqrt{W/\nu}$ where $Z$ is standard normal and $W$ is $\chi^2_\nu$. $\endgroup$ – Xi'an Mar 24 '17 at 8:48
  • $\begingroup$ @JohnK Central. I'll edit the question to make it more clear. @Xi'an Can this be applied to the non central distribution? Your comment has confused me somewhat. I understand if v is large enough the distribution converges to a normal. But how do you get it by not giving v a value? $\endgroup$ – Humza Ali Mar 24 '17 at 9:24
  • $\begingroup$ you could try to use the characteristic function $\endgroup$ – kjetil b halvorsen Jun 10 '17 at 19:27
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There are various ways to find the moments of the T-distribution, but the simplest method is to use the mixture representation using the normal distribution. If $T$ has a Student's T distribution with $\varphi$ degrees-of-freedom then we can write it via the mixture $T|\lambda \sim \text{N}(0, \tfrac{1}{\lambda})$ with $\lambda \sim \text{Ga}(\tfrac{\varphi}{2}, \tfrac{\varphi}{2})$ (i.e., as a mixture of normal distributions where the precision is gamma distributed). Using this mixture representation the density function for the Student's T distribution can be written as the integral:

$$\begin{equation} \begin {aligned} \text{St}(t|\varphi) &= \int \limits_0^\infty \text{N}(t|0,\tfrac{1}{\lambda}) \text{Ga}(\lambda|\tfrac{\varphi}{2},\tfrac{\varphi}{2}) \ d \lambda. \end{aligned} \end{equation}$$

Using this mixture representation, the raw moments of $T$ can be obtained via the law of iterated expectation, using the known moments of the normal distribution. The conditional moments are:

$$\mathbb{E}(T^k|\lambda) = \int \limits_{-\infty}^\infty t^k \text{ N}(t|0,\tfrac{1}{\lambda}) \text{ } dt = \begin{cases} 0 & & \text{if } k \text{ is odd}, \\[6pt] \frac{k!}{2^{k/2}(k/2)!} \lambda^{-k/2} & & \text{if } k \text{ is even}. \\[6pt] \end{cases}$$

For values $k \geqslant \varphi$ the moments of the T distribution do not exist. For odd values $0<k<\varphi$ the moments are zero and for even values $0<k<\varphi$ the moments are:

$$\begin{equation} \begin {aligned} \mathbb{E}(T^k) = \mathbb{E}( \mathbb{E}(T^k | \lambda ) ) &= \int \limits_0^\infty \frac{k!}{2^{k/2}(k/2)!} \lambda^{-k/2} \text{ Ga}(\lambda|\tfrac{\varphi}{2},\tfrac{\varphi}{2}) \text{ } d\lambda \\[6pt] &= \int \limits_0^\infty \frac{k!}{2^{k/2}(k/2)!} \lambda^{-k/2} \cdot \frac{\varphi^{\varphi/2}}{2^{\varphi/2} \Gamma(\tfrac{\varphi}{2})} \lambda^{\varphi/2-1} \exp \Big( - \frac{\varphi}{2} \lambda \Big) \text{ } d\lambda \\[6pt] &= \frac{k!}{2^k (k/2)!} \cdot \frac{\Gamma(\tfrac{\varphi-k}{2})}{\Gamma(\tfrac{\varphi}{2})} \cdot \varphi^{k/2} \\ &\quad \quad \quad \quad \times \int \limits_0^\infty \frac{\varphi^{(\varphi-k)/2}}{2^{(\varphi-k)/2} \Gamma(\tfrac{\varphi-k}{2})} \lambda^{(\varphi-k)/2-1} \exp \Big( -\frac{\varphi}{2} \lambda \Big) \text{ } d\lambda \\[6pt] &= \frac{k!}{2^k (k/2)!} \cdot \frac{\Gamma(\tfrac{\varphi-k}{2})}{\Gamma(\tfrac{\varphi}{2})} \cdot \varphi^{k/2} \int \limits_0^\infty \text{Ga}(\lambda|\tfrac{\varphi-k}{2},\tfrac{\varphi}{2}) \text{ } d\lambda \\[6pt] &= \frac{k!}{2^k (k/2)!} \cdot \frac{\Gamma(\tfrac{\varphi-k}{2})}{\Gamma(\tfrac{\varphi}{2})} \cdot \varphi^{k/2} \\[6pt] &= \frac{\Gamma(\tfrac{k+1}{2})}{\sqrt{\pi}} \cdot \frac{\Gamma(\tfrac{\varphi-k}{2})}{\Gamma(\tfrac{\varphi}{2})} \cdot \varphi^{k/2}. \\[6pt] &= \frac{\Gamma(\tfrac{k+1}{2})}{\sqrt{\pi}} \cdot \frac{\varphi^{k/2}}{\prod_{i=1}^{k/2} (\tfrac{\varphi}{2}-i)}. \\[6pt] \end{aligned} \end{equation}$$

Application of this formula for the even moments yields:

$$\begin{equation} \begin{aligned} \mathbb{E}(T^2) &= \frac{\varphi}{\varphi-2} & & & \text{for } \varphi > 2, \\[6pt] \mathbb{E}(T^4) &= \frac{3 \varphi^2}{(\varphi-2) (\varphi-4)} & & & \text{for } \varphi > 4, \\[6pt] \mathbb{E}(T^6) &= \frac{15 \varphi^3}{(\varphi-2) (\varphi-4) (\varphi-6)} & & & \text{for } \varphi > 6. \\[6pt] \end{aligned} \end{equation}$$

You can now obtain the kurtosis, etc., via algebraic manipulation of the raw moments. In particular, the kurtosis is:

$$\begin{equation} \begin{aligned} \mathbb{Kurt}(T) = \frac{\mathbb{E}(T^4)}{\mathbb{E}(T^2)^2} &= \frac{3 \varphi^2}{(\varphi-2) (\varphi-4)} \Big/ \Big( \frac{\varphi}{\varphi-2} \Big)^2 \\[6pt] &= \frac{3 (\varphi-2)}{\varphi-4} \\[6pt] &= \frac{3 \varphi - 6}{\varphi-4} \\[6pt] &= \frac{3 \varphi - 12}{\varphi-4} + \frac{6}{\varphi-4} \\[6pt] &= 3 + \frac{6}{\varphi-4}. \\[6pt] \end{aligned} \end{equation}$$

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