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I am looking for a resource where i can find derivation of the linear combination of multivariate t distribution. Does anyone here know any good site or place (s)he can point me to? I am trying to see if the linear combination of multivariate t distribution will give a multivariate t distribution. In other words, what is the distribution of the linear combination of two or more multivariate t distribution?

Does anyone here have any idea on this?

thanks.

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    $\begingroup$ (i) why is 'algorithms' one of the tags? (ii) are the components independent? $\endgroup$
    – Glen_b
    Dec 29 '13 at 16:22
  • $\begingroup$ Yes, the components are independent. Sorry about the algorithm tag. $\endgroup$
    – elizwet
    Dec 29 '13 at 17:03
  • $\begingroup$ One thing I should have asked you to clarify before. When you say 'linear combination of distribution' I have assumed you meant to take a linear combination of the random variables (where 'sum' is really a convolution of pdfs/probability functions in the independent case), not of the distribution functions (i.e. a mixture model). $\endgroup$
    – Glen_b
    Dec 30 '13 at 11:41
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I am trying to see if the linear combination of multivariate t distribution will give a multivariate t distribution.

In general, no, this is not the case, even with univariate t's (see here and here for example; note that the difference of two t-random variables is the sum of two t-random variables, but with the second component having its mean that of the original random variable multiplied by -1)

In some very particular cases, yes. Consider:

(i) the limiting case of infinite degrees of freedom, linear combinations of multivariate normals are multivariate normal;

(ii) if the component t-variables are perfectly dependent their sums will be multivariate-t;

(iii) in the univariate case, sums of independent Cauchy random variables will be Cauchy. I haven't checked, but this may well extend more to subsets of the multivariate case than vectors of independent Cauchy (and the perfectly-dependent case mentioned above);

(iv) in the limit of very large numbers of components, where none of the components dominates variance-wise (that is, where the coefficient of each component times the variance of that component doesn't become too large), you may be able to invoke a version of the central limit theorem.


In the case where the weights on the components are equal (effectively converting it to a scaled sum) and you're dealing with standard t (rather than ones with general means and variances), this paper has some information. Extending it to the case of a general mean is straightforward but it doesn't deal with the general case of arbitrary scales, or equivalently arbitrary linear combinations.

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  • $\begingroup$ I know that multivariate normals are normals when u add the independent components. How comes the linear combination of multivariate t are not multivariate t? Multivariate t are instances of the multivariate norma( i mean combinations). $\endgroup$
    – elizwet
    Dec 29 '13 at 17:08
  • $\begingroup$ What do you mean by the sentence "multivariate t are instances of the multivariate normal"? $\endgroup$
    – Glen_b
    Dec 29 '13 at 17:30
  • $\begingroup$ Sorry, i meant to say, the multivariate t distribution is a family of continuous probability distributions with different degree of freedom. Say when df=1, you get Cauchy.So maybe i should have said with different df, you can get instances of the multivariate t distribution. $\endgroup$
    – elizwet
    Dec 29 '13 at 17:44
  • $\begingroup$ Yes, but now how does this relate to what you just asked me in comments a few minutes ago? The fact that you know a limiting case has a certain property doesn't imply that the general case shares that property. Your question there implies that you expect it should be the case. What is the basis for such an expectation? $\endgroup$
    – Glen_b
    Dec 29 '13 at 17:47
  • $\begingroup$ I base it on the fact that the big picture(here multivariate t) will inherit the charactersitics of the instances( Say cauchy). this is just a feeling, it maybe be a total rusbish. So i just came here to seek advice from experts like you so that i can be enlighten. So if i have say two independent multivariate t distribution, when i add them up, will the results give me a multivariate t distribution as in the case of multivariate Gaussian distribution? $\endgroup$
    – elizwet
    Dec 29 '13 at 18:07
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Please have a look at Walker, Glenn A., and John G. Saw. "The distribution of linear combinations of t-variables." Journal of the American Statistical Association 73.364 (1978): 876-878.

The resulting PDF is described as a weighted sum of student-t distribution, and the paper shows how to obtain the weight. The author started from observing for odd degrees of freedom the characteristic function of a student-t r.v. is expressible in closed form, i.e. proportional to the modified Bessel function of the third kind. It seems to me that the paper proposes the solution for all odd t-distribution cases, and no even degree-of-freedoms should be involved.

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    $\begingroup$ Welcome to the site. At present this is more of a comment than an answer. You could expand it, perhaps by giving a summary of the information in the paper, or we can convert it into a comment for you. $\endgroup$ Nov 21 '15 at 19:59
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P15 of Multivariate T distribution and their application (Kotz and Nadarajah) says " If X has the p-variate t distribution with degrees of freedom v, mean vector p, and correlation matrix R, then, for any nonsingular scalar matrix C and for any a, CX + a has the p-variate t distribution with degrees of freedom v, mean vector Cp+a, and correlation matrix CRC'. This result is of importance in applications and is similar to the corresponding result for the multivariate normal distribution. "

edit:

In a univariate language, e.g., considering $aT_1+bT_2$ where $T_i$ is t distributed with mean $m_1,m_2$, scale (in same order as variance) $S_1,S_2$ and with the same degree of freedom. Then define $T:=[T1, T2]'$ which is a bivariate T distribution with zero covariance by stacking the $T_1$ and $T_2$ into a vector. Then $aT_1+bT_2=[a, b]T$ . Applying above conclusion, $aT_1+bT_2$ is univariate t distributed with the same degree of freedom, whose mean is $am_1+bm_2$ and scale is $a^2S_1+b^2S_2$.

An R simulation that supports this theory is:

require('mas3321')
n=10000
sample=c()
mean1=.6
mean2=1.2
scale1=.5
scale2=1
p1=10
p2=20
samples_combt=p1*rgt(n, 10, mean1, scale1)+p2*rgt(n, 10,mean2 , scale2)
hist(samples_combt,probability = T)    
mean_comb= p1*mean1+p2*mean2
scale_comb=p1^2*scale1+p2^2*scale2
curve(dgt(x, 10,mean_comb, scale_comb), add=TRUE)
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If $X$ follows a multivariate t-distribution, then any linear combination of $X$ also follows a multivariate t-distribution with the same degrees of freedom:

$$ X \sim t(\mu, \Sigma, \nu) \quad \Rightarrow \quad Y = AX + b \sim t(A\mu + b, A\Sigma A^\mathrm{T}, \nu) \; . $$

Thus, the intended combination only works, if the two multivariate t-distributions have the same dimensions and degrees of freedom. Let us assume that

$$ \begin{split} X_1 &\sim t(\mu_1, \Sigma_1, \nu) \\ X_2 &\sim t(\mu_2, \Sigma_2, \nu) \end{split} $$

where $X_1$ and $X_2$ are independent $n \times 1$ random vectors. If that is the case, we have:

$$ X = \left[ \begin{array}{c} X_1 \\ X_2 \end{array} \right] \sim t\left( \left[ \begin{array}{c} \mu_1 \\ \mu_2 \end{array} \right], \; \left[ \begin{array}{cc} \Sigma_1 & 0_{nn} \\ 0_{nn} & \Sigma_2 \end{array} \right], \; \nu \right) \; . $$

The random variable you seem to have in mind is probably this:

$$ Y = c_1 X_1 + c_2 X_2 \; . $$

Note that $Y$ can be emulated from $X$ by specifying an appropriate linear combination:

$$ A = \left[ \begin{array}{c} c_1 I_n & c_2 I_n \end{array} \right], \; b = 0_{n} \quad \Rightarrow \quad Y = AX + b = c_1 X_1 + c_2 X_2 \; . $$

Thus, we can apply the linear transformation theorem from above:

$$ Y \sim t\left( \left[ \begin{array}{c} c_1 I_n & c_2 I_n \end{array} \right] \left[ \begin{array}{c} \mu_1 \\ \mu_2 \end{array} \right] + 0_{n}, \; \left[ \begin{array}{c} c_1 I_n & c_2 I_n \end{array} \right] \left[ \begin{array}{cc} \Sigma_1 & 0_{nn} \\ 0_{nn} & \Sigma_2 \end{array} \right] \left[ \begin{array}{c} c_1 I_n & c_2 I_n \end{array} \right]^\mathrm{T}, \; \nu \right) \; . $$

This gives:

$$ Y \sim t\left( c_1 \mu_1 + c_2 \mu_2, \; c_1^2 \Sigma_1 + c_2^2 \Sigma_2, \; \nu \right) \; . $$

Fun Fact: The above theorem can also be used to prove the relationship between the multivariate t-distribution and the F-distribution.

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Student-t distribution is a special case of the Generalised Hyperbolic Distribution which is closed under affine transform according to wikipedia page (all linear transforms are affine transforms). Hence I would think all linear transformations of student-t random variables (with same degree of freedom) are student-t distributed.

I believe if $$X \sim t_d(\nu, \mathbf{\mu}, \Sigma)$$ then $$\mathbf{w}^T X + c \sim t_1(\nu, \mathbf{w}^T \mathbf{\mu} + c, \mathbf{w}^T \Sigma \mathbf{w})$$

This is just my understanding after reading the below references and adds to answer posted by @user31575.

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

Also in "Quantitative risk management: Concepts, techniques and tools" section 2.3.1 equation 2.31

Also referenced in this paper Hu, W. and Kercheval, A.N., 2010. Portfolio optimization for student t and skewed t returns.

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  • $\begingroup$ hence I think the addition of two uncorrelated student-t random variables (which isn't the same as independent) is also student-t distributed $\endgroup$ Dec 30 '19 at 21:20

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