Question
Show that for the $t$ distribution with $v>4$ degrees of freedom(a) $\mu_{4}=\frac{3 v^{2}}{(v-2)(v-4)}$;(b) $\alpha_{4}=3+\frac{6}{v-4}$.(Hint: Make the substitution $1+\frac{t^{2}}{v}=\frac{1}{u} .$ )
Step 1
Rearranging this gives us \( t^2 = v\left(\frac{1}{u} - 1\right) = \frac{v - u}{u} \). Show more…
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