91影视

The derivative of the CDF is the probability density function $f\left(x\right)$, abbreviated PDF if it exists. A distribution function$F_x$ describes each random variableX.

We are given that X has a Gamma distribution with parameters $tand位$. Let's find the CDF of cX. We have that

$$\begin{gathered} F_{cX}\left(z\right)=P(cX鈮�z) \\ =P(X鈮�\frac{z}{c})=F_X\left(\frac{z}{c}\right) \end{gathered}$$

Hence,

$f_{cX}\left(z\right)=\frac{d}{dz}{F}_X\left(\frac{z}{c}\right)=f_X\left(\frac{z}{c}\right).\frac{1}{c}$

which implies that,

$$\begin{gathered} f_{cX}\left(z\right)=\frac{{位}^t}{蟿\left(t\right)}{\left(\frac{z}{c}\right)}^{t-1}{e}^{-位\frac{z}{c}}.\frac{1}{c} \\ =\frac{{\left({\displaystyle \frac{位}{c}}\right)}^t}{蟿\left(t\right)}{z}^{t-1}{e}^{-\frac{位}{c}z} \end{gathered}$$

So, we see that$cX~饾殴(t,\frac{位}{c})$

Take any z > 0 and define $Z:\frac{1}{2位}{X^2}_{2n}$, we have that

$$\begin{gathered} F_z\left(z\right)=P(Z鈮�z) \\ =P(\frac{1}{2位}{X^2}_{2n}鈮�z) \\ =P({X^2}_{2n}鈮�2位z) \\ ={F_{x^2}}_{2n}\left(2位z\right) \end{gathered}$$

Hence we get that

$$\begin{gathered} f_Z\left(z\right)=\frac{d}{dz}{F}_z\left(z\right) \\ =\frac{d}{dz}{F}_{{x^2}_{2n}}\left(2位z\right) \\ =\frac{{位}^n}{蟿\left(n\right)}{z}^{n-1}{e}^{-位z} \end{gathered}$$

A chi-squared random variable with 2n degrees of freedom can be regarded as being the sum of n independent chi-square random variables each with 2 degrees of freedom (which for Example is equivalent to an exponential random variable with a parameter ). Hence Proposition ${X^2}_{2n}$is a gamma random variable with parameters $(n,1/2)$and the results now follow from part (a)