# Integration of student's T PDF

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: $$\int_{-\infty}^\infty t(x,\nu)t(x+a,\nu)dx = \frac{1}{\sqrt2}T\left(\frac{a}{\sqrt2},2\nu\right)$$

• Are you sure this property holds at all? – Christoph Hanck Sep 28 '16 at 5:01
• 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. – whuber Sep 29 '16 at 17:05
• @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. – Kevin Nowaczyk Sep 29 '16 at 18:12
• math.stackexchange.com/questions/1946862/… – Kevin Nowaczyk Sep 29 '16 at 18:39

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.