Question

Question: Prove the limit formula Eq. (8.4.6).

Short answer

The limit formula of the equation 8.4.6 is \(\mathop {\lim }\limits_{m \to \infty } \frac{{\left| \!{\overline {\, {\left( {m + \frac{1}{2}} \right)} \,}} \right. }}{{\left| \!{\overline {\, {m{{\left( m \right)}^{\frac{1}{2}}}} \,}} \right. }} = 1\)

Step-by-step solution

Given information

The pdf of the random variable X from t-distribution with n degrees of freedom is

\(f\left( x \right) = \frac{{\left| \!{\overline {\, {\left[ {\left( {\frac{{n + 1}}{2}} \right)} \right]} \,}} \right. }}{{{{\left( {n\pi } \right)}^{\frac{1}{2}}}\left| \!{\overline {\, {\left( {\frac{n}{2}} \right)} \,}} \right. }}{\left( {1 + \frac{{{x^2}}}{n}} \right)^{ - \left( {\frac{{n + 1}}{2}} \right)}}\)

It is needed to prove that

\(\mathop {\lim }\limits_{m \to \infty } \frac{{\left| \!{\overline {\, {\left( {m + \frac{1}{2}} \right)} \,}} \right. }}{{\left| \!{\overline {\, {m{{\left( m \right)}^{\frac{1}{2}}}} \,}} \right. }} = 1\)

Proof of \(\mathop {\lim }\limits_{m \to \infty } \frac{{\left| \!{\overline {\, {\left( {m + \frac{1}{2}} \right)} \,}} \right. }}{{\left| \!{\overline {\, {m{{\left( m \right)}^{\frac{1}{2}}}} \,}} \right. }} = 1\) 

From the limit formula of the equation 8.4.2 it can be shown that

\(\begin{align}\mathop {\lim }\limits_{m \to \infty } \frac{{\Gamma \left( {m + \frac{1}{2}} \right)}}{{\Gamma \left( m \right){m^{\frac{1}{2}}}}} &= \mathop {\lim }\limits_{m \to \infty } \frac{{{{\left( {2\pi } \right)}^{\frac{1}{2}}}{{\left( {m + \frac{1}{2}} \right)}^{\left( {m + \frac{1}{2}} \right) - \frac{1}{2}}}{e^{ - \left( {m + \frac{1}{2}} \right)}}}}{{{{\left( {2\pi } \right)}^{\frac{1}{2}}}{m^{\frac{{m - 1}}{2}}}{e^{ - m}}{m^{\frac{1}{2}}}}}\\ &= \mathop {\lim }\limits_{m \to \infty } \frac{{{{\left( {m + \frac{1}{2}} \right)}^m}}}{{{m^m}}}{e^{ - \frac{1}{2}}}\\ &= \mathop {\lim }\limits_{m \to \infty } {\left( {1 + \frac{{\frac{1}{2}}}{m}} \right)^m}{e^{ - \frac{1}{2}}}\\ &= {e^{\frac{1}{2}}}{e^{ - \frac{1}{2}}}\\ &= 1\end{align}\)

Hence the proof.