问题及解答

Evaluate the following:

• (a) $\Gamma(6)$
• (b) $\Gamma(\frac{5}{2})$
• (c) $F(4;5)$  (the incomplete gamma function)
• (d) $F(5;4)$
• (e) $F(0;4)$
   

(a) Recall that we have the formula about gamma function (请参见问题714)

\[
\Gamma(x+1)=x\cdot\Gamma(x),\quad\forall\ x > 0.
\]

Hence, for positive integer $n$, we have

\[
\begin{split}
\Gamma(n+1)=&n\cdot\Gamma(n)\\
&=n(n-1)\cdot\Gamma(n-1)\\
&=n(n-1)(n-2)\cdot\Gamma(n-2)\\
&\vdots\\
&=n(n-1)(n-2)\cdots 3\cdot 2\Gamma(2)\\
&=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\Gamma(1)\\
\end{split}
\]

where

\[\Gamma(1)=\int_{0}^{\infty}t^{1-1}e^{-t}dt=\int_{0}^{\infty}e^{-t}dt=-e^{-t}\biggr|_{0}^{\infty}=1\]

Therefore,

\[
\Gamma(n+1)=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1=n!
\]

So, $\Gamma(6)=5!=5\times 4\times 3\times 2\times 1=120$.

(b)

\[
\begin{split}
\Gamma(\frac{5}{2})&=\frac{3}{2}\cdot\Gamma(\frac{3}{2})\\
&=\frac{3}{2}\cdot\frac{1}{2}\cdot\Gamma(\frac{1}{2})\\
&=\frac{3}{2}\cdot\frac{1}{2}\cdot\sqrt{\pi}\\
&=\frac{3}{4}\sqrt{\pi}
\end{split}
\]

(c)

Recall that $F(x;\alpha)$ is the cdf of a standard gamma rv $X$, 

\[
F(x;\alpha)=\int_{0}^{x}\frac{t^{\alpha-1}e^{-t}}{\Gamma(\alpha)}dt=\frac{1}{\Gamma(\alpha)}\cdot\int_{0}^{x}t^{\alpha-1}e^{-t}dt
\]

Then,

\[
\begin{split}
F(4;5)&=\frac{1}{\Gamma(5)}\cdot\int_{0}^{4}t^{5-1}e^{-t}dt\\
&=\frac{1}{4!}\cdot\int_{0}^{4}t^4 e^{-t}dt\\
&=\frac{1}{24}\cdot\int_{0}^{4}t^4 e^{-t}dt
\end{split}
\]

Now we calculate the integral $\int_{0}^{4}t^4 e^{-t}dt$.

\[
\begin{split}
\int_{0}^{4}t^4 e^{-t}dt&=-\int_{0}^{4}t^4 de^{-t}=(-1)\biggl[t^4 e^{-t}\biggr|_{0}^{4}-\int_{0}^{4}e^{-t}dt^4\biggr]\\
&=\int_{0}^{4}e^{-t}\cdot 4t^3 dt-4^4 e^{-4}=-\int_{0}^{4}4t^3 de^{-t}-4^4 e^{-4}\\
&=-\biggl[4t^3 e^{-t}\biggr|_{0}^{4}-\int_{0}^{4}e^{-t}d(4t^3)\biggr]-4^4 e^{-4}\\
&=\int_{0}^{4}12e^{-t}t^2 dt-4\cdot 4^3 e^{-4}-4^4 e^{-4}=-12\int_{0}^{4}t^2 de^{-t}-2\cdot 4^4 e^{-4}\\
&=-12\biggl[t^2 e^{-t}\biggr|_{0}^{4}-\int_{0}^{4}e^{-t}dt^2\biggr]-2\cdot 4^4 e^{-4}\\
&=12\int_{0}^{4}e^{-t}\cdot 2tdt-12\cdot 4^2 e^{-4}-2\cdot 4^4 e^{-4}\\
&=-24\int_{0}^{4}tde^{-t}-(3\cdot 4^3+2\cdot 4^4)e^{-4}\\
&=-24\biggl[te^{-t}\biggr|_{0}^{4}-\int_{0}^{4}e^{-t}dt\biggr]-11\cdot 4^3 e^{-4}\\
&=24\int_{0}^{4}e^{-t}dt-24\cdot 4e^{-4}-11\cdot 4^3 e^{-4}\\
&=24\cdot(-e^{-t})\biggr|_{0}^{4}-(96+11\cdot 4^3)e^{-4}\\
&=24(1-e^{-4})-800e^{-4}=24-824e^{-4}.
\end{split}
\]

Hence,

\[
F(4;5)=\frac{1}{24}\cdot(24-824e^{-4})\approx 0.37116307
\]

Remark:

We can also get the value of $F(4;5)$ by checking Table A.4. It is $0.371$.

(d)

From Table A.4, we have

\[
F(5;4)\approx .735
\]

(e)

\[
F(0;4)=\frac{1}{\Gamma(4)}\cdot\int_{0}^{0}t^{4-1}e^{-t}dt=0
\]