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The standard normal distribution has the property that

$$\int_{-\infty}^\infty \phi(x)\phi(x+a)dx = \frac{1}{\sqrt2}\phi\left(\frac{a}{\sqrt2}\right)$$

How would I go about proving that the same property holds for the Student's T distribution? I've been messing around with integration by parts and u substitution to no avail. The key is solving the following. $$\int_{-\infty}^\infty \left(1+\frac{(u+a)^2}{\nu}\right)^{-\frac{\nu+1}{2}}\left(1+\frac{u^2}{\nu}\right)^{-\frac{\nu+1}{2}}du$$

Edit: some values of critical q values at k=2:

|df    |q (.95)  |q(.99)   |
|5     |3.635350 |5.702312 |
|6     |3.460456 |5.243097 |
|7     |3.344085 |4.949044 |
|8     |3.261182 |4.745232 |

Edit 2: After thinking about how the studentized range distribution comes into being, I think the degrees of freedom will be different in the left and right of the equality. This isn't an issue when using the standard normal.

I think the thing to prove is: $$\int_{-\infty}^\infty t(x,\nu)t(x+a,\nu)dx = \frac{1}{\sqrt2}T\left(\frac{a}{\sqrt2},2\nu\right)$$

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    $\begingroup$ Are you sure this property holds at all? $\endgroup$ – Christoph Hanck Sep 28 '16 at 5:01
  • $\begingroup$ It would be more constructive to ask how to evaluate that integral rather than to speculate about the answer, which is wrong, as @glen_b has indicated (and Christoph Hanck and I have both intimated, too). When you ask us to prove something that is false, exactly what would constitute a good answer? One counterexample? You won't make much progress that way. $\endgroup$ – whuber Sep 29 '16 at 17:05
  • $\begingroup$ @whuber - True...I originally thought that was the question, but it turns out it isn't. The real question is, if that integral does not evaluate similarly, how does the final expression come about. I assumed there was some polynomial or hypergeometric property out there that took care of it. Which it why I edited the question, to show that the original question is not right. $\endgroup$ – Kevin Nowaczyk Sep 29 '16 at 18:12
  • $\begingroup$ math.stackexchange.com/questions/1946862/… $\endgroup$ – Kevin Nowaczyk Sep 29 '16 at 18:39
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It holds for the normal for the (fairly obvious) reason that the sum of two quadratics is quadratic; completing the square and recognizing a density integrates to 1 will then give the result.

There's nothing obvious that would seem to suggest it should hold for the t. ... why do you think it does?

Indeed, if it did hold, it would suggest that the sum of two independent $t_\nu$ random variates would have a t-distribution (since it's of almost the same form as the convolution integral) -- but that is not the case, so my expectation is that it generally doesn't hold.

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  • $\begingroup$ This comes about from the Studentized Range Distribution. If this distribution is evaluated with infinite degrees of freedom and two groups, it evaluates as the standard normal expression given above. When the degrees of freedom are not infinite, it evaluates as the Student's T version. The Studentized Range Distribution is constructed by the integral above. I am also dubious that this would have been the case, but the numbers match. $\endgroup$ – Kevin Nowaczyk Sep 28 '16 at 11:44
  • $\begingroup$ The Residue Theorem indicates we should expect the result to be true for 1 DF and that something a little bit like it will hold for any positive odd DF. $\endgroup$ – whuber Sep 28 '16 at 13:53
  • $\begingroup$ Can you clarify what numbers match what? $\endgroup$ – Glen_b -Reinstate Monica Sep 28 '16 at 14:12
  • $\begingroup$ @whuber yeah it's going to be true for 1 df because an average of iid cauchy's is cauchy (which was enough for me to say "not in general" rather than "not true at all") $\endgroup$ – Glen_b -Reinstate Monica Sep 28 '16 at 14:13
  • $\begingroup$ I've added some critical q values. These are the values of q such that F(q, df,k) = .95 or .99. They match the table values and are calculated from 2*T(q/sqrt(2),df) - 1. The wikipedia article on critical q values also mentions that one can determine if the square-root of two is or is not included by comparing the value to a t-test. $\endgroup$ – Kevin Nowaczyk Sep 28 '16 at 15:13

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