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When DF = 1, t distribution is just Cauchy distribution whose mean does not exist.

$E(T) = E(\frac{Z_1}{\sqrt{Z_2^2}}) = E(\frac{Z_1}{Z_2}) = E(Z_1)E(\frac{1}{Z_2})$ where the second expectation converges to infinity because $E(Z_2) = 0$

Similarly when DF = 2, I want the second moment (Suppose V follows Chi Square distribution):

$E(T^2) = E(\frac{Z_1^2}{(\sqrt{V_2/2})^2}) = E(\frac{2Z_1^2}{V_2}) = 2E(Z_1^2)\frac{1}{E(V_2)} = 1$ since $E(Z_1^2) = 1$ and $E(V_2) = 2$

So I'm wondering what's wrong with my equation above? When DF = 2, the second moment shouldn't exist.

Similarly, when DF = 3, I want to calculate the variance, from the formula I have $3/(3 - 2) = 3$

However, I can also show that:

$Var(T) = E(T^2) - E(T)^2= E(\frac{Z_1^2}{(\sqrt{V_3/3})^2}) - 0 = E(\frac{3Z_1^2}{V_3}) = 3E(Z_1^2)\frac{1}{E(V_3)} = 1$

My steps must be wrong somewhere, but I just don't understand it.

I know I can show second moment doesn't exist by plugging $X^2$ to the PDF.

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    $\begingroup$ Add the self-study tag. $\endgroup$ – Michael R. Chernick Nov 22 '19 at 4:03
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    $\begingroup$ Hint: $E\left(\frac{2Z^2}{V}\right) = 2E(Z^2)E\left(\frac{1}{V}\right) \neq 2E(Z^2)\frac{1}{E(V)}$ $\endgroup$ – knrumsey Nov 22 '19 at 4:05
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    $\begingroup$ What about considering the integral for the second non-central moment of the $t_2$ distribution and looking at its equivalent expression at $\pm\infty$? $\endgroup$ – Xi'an Nov 22 '19 at 8:35

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