Show that t distribution variance doesn't exist when df = 2

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.

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