Distribution of sum of squares of normal distribution

Question

Given X1,...X7 are independent and identically distributed from a normal distribution with N(1,2)

Denoting: $$\bar{X} =\frac17\sum_{i=1}^7(X_i)$$ What is the exact distribution of V below? $$V =\sum_{i=1}^7(X_i-\bar{X})^2 $$

I have derived that $$Y =\frac{\sum_{i=1}^7(X_i-\bar{X})^2}{2} $$ Y is a chi-square distribution with degree of freedom=6.

However, i am unsure of V, which does not have the factor 1/2

Answer 1

If the distribution of $Y$ is $\chi^2_6=\Gamma(k=3,\theta=2)$ with the following pdf $$ f_Y(y)=\frac{1}{\Gamma(k)\theta^k}y^{k-1}e^{-x/\theta}=\frac{1}{2^4}y^{2}e^{-x/2}, \quad y>0 $$ then $V=2Y$ follows Gamma distribution $\Gamma(k=3,\theta=4)$ with pdf $$ f_V(v) = \frac{1}{\Gamma(k)\theta^k}y^{k-1}e^{-x/\theta}=\frac{1}{2\cdot 4^3}v^{2}e^{-v/4}, \quad v>0. $$ Look here for the property

If $X\sim\chi^2(\nu)$ and $c>0$ then $cX\sim\Gamma(k=\nu/2,\theta=2c)$.