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... and why ?

Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:

  • $E(X_1-X_2)=\mu_1-\mu_2$
  • $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$

Now let's say $X_1$, $X_2$ are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?

This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.

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    $\begingroup$ this is related question which might be of interest. $\endgroup$
    – mpiktas
    May 16, 2011 at 14:46
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    $\begingroup$ Google the Satterthwaite t-test, the CABF t-test (Cochran's approximation to the Behrens-Fisher), and the Behrens-Fisher problem. $\endgroup$
    – whuber ♦
    May 16, 2011 at 15:51
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    $\begingroup$ For the special case where the degrees of freedom is 1 (the Cauchy distribution) the answer to the original question is 1. The sum (or difference) of two independent Cauchy distributed random variables is Cauchy with scale parameter $2$, but then again, the Cauchy distribution does not even have a mean value. $\endgroup$
    – NRH
    May 16, 2011 at 17:20
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    $\begingroup$ You need to check the Behrens–Fisher distribution $\endgroup$
    – Wis
    Mar 6, 2016 at 19:53

2 Answers 2

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The sum of two independent t-distributed random variables is not t-distributed. Hence you cannot talk about degrees of freedom of this distribution, since the resulting distribution does not have any degrees of freedom in a sense that t-distribution has.

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  • $\begingroup$ @mpiktas: Dumb question. If the t-distribution with n-1 df can be derived from the sum of n indepent normal distributions (see wikipedia) and given df high enough so that the t-distribution approximates the normal distribution, doesn't derive from that that the sum of t-distributions is again a t-distribution ? $\endgroup$
    – steffen
    May 16, 2011 at 14:15
  • $\begingroup$ @mpiktas: What about the test-statistic of the t-test, which seems to be derived from the difference of two t-distributions ? $\endgroup$
    – steffen
    May 16, 2011 at 14:17
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    $\begingroup$ @steffen, no. It will be approximately normal, since you will add two approximately normal distributed normal variables. t-distribution with high df is approximately normal, but approximately normal is not necessarily t-distribution with high df. $\endgroup$
    – mpiktas
    May 16, 2011 at 14:19
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    $\begingroup$ @steffen, t-test statistic is derived from the difference of two normals not two t-distributions. Note that definition of t distribution is a fraction of normal and square root of chi-square. $\endgroup$
    – mpiktas
    May 16, 2011 at 14:21
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    $\begingroup$ @steffen, I often say to my students there are no stupid questions, only stupid people who do not ask any questions. I am not a very popular teacher I should add :) $\endgroup$
    – mpiktas
    May 16, 2011 at 14:35
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Agree the answers above, the difference of two independent t-distributed random variables are not t distributed. But I want to add some ways of calculating this.

  1. The easiest way of calculating this is using a Monte Carlo method. In R, for example, you random sample 100,000 numbers from the first t distribution, then you random sample another 100,000 numbers from the second t distribution. You let the first set of 100,000 numbers minus the second set of 100,000 numbers. The obtained 100,000 new numbers are the random samples from the distribution of the difference between the two distribution. You can calculate the mean and variance by simply using mean() and var().

    1. This is called Behrens–Fisher distribution. You can refer to the Wiki page: https://en.wikipedia.org/wiki/Behrens%E2%80%93Fisher_distribution. The CI given by this distribution is called "fiducial interval", this is not a CI.

    2. Numerical integration might work. This is continued as the bullet point 2. You might refer to the Section 2.5.2 in Bayesian Inference in Statistical Analysis by Box, George E. P., Tiao, George C. It has the detailed steps of integration, and how this is approximated to be a Behrens–Fisher distribution.

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    $\begingroup$ It seems to me that the Behrens-Fisher distribution applies where the variance of the two t-distributions are not equal. Can the same be said if the variance of the two distributions IS equal? $\endgroup$
    – Ian Sudbery
    Mar 9, 2017 at 22:46
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    $\begingroup$ Sorry, pressed enter two early? To continue... For example say we have two normal distributions of equal but unknown variance, but different means. We draw two samples from each of these distributions. The difference of means between the two samples from the same distribution will follow a t-distribution, but what is the distribution of the difference of the differences. $\endgroup$
    – Ian Sudbery
    Mar 9, 2017 at 22:53

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