Alternate method to find distribution of function of X Let $X_1,X_2 ,... X_n ;(n\geq1)$ be a random sample from $Exp(1)$. Then the distribution of $2n\bar X$ is ?
$(A) Exp\bigg(\dfrac{1}{2}\bigg)$
$(B) Exp(2n)$
$(C)\ \ \chi^2_{(n)}$
$(D)\ \ \chi^2_{(2n)}$
I solved this problem using Mgf :
Let $Y=\sum X_i$
$Y\sim G(1,n)$ 
$M_Y(t)=(1-t)^{-n}$
$M_{2Y}(t)=(1-2t)^\frac{-2n}{2} \sim \chi^2_{(2n)}$ 
Correct me in the above steps and please tell me any alternate method(s) which cab be used in this problem or any shortcut. This question came in 1 mark so i think there is some shortcut here. (Poor English sorry ).
Gamma distribution :If

$Y\sim G(a,\lambda)=$ $\dfrac{\alpha^{
 \lambda}}{\Gamma{(\lambda)}}e^{-ax}x^{\lambda-1} \ \ ; \ \ (a>0) ,(\lambda>0), (x>0)$

(Please try to use this notation for gamma density i just started learning these things so it can be a case i might get confused)
.
 A: It is more common to use $\mathsf{Gamma}(\lambda,a)$ or $\Gamma(\lambda,a)$ in order to denote the gamma distribution (parameters switched in order).
It is well known (and handsome to know) that - if $X_1,\dots, X_n$ are independent with $X_i\sim\Gamma(\lambda_i,a)$:$$\sum_{i=1}^nX_i\sim\Gamma\left(\sum_{i=1}^n\lambda_i,a\right)$$
In your question we have $X_i\sim\Gamma(1,1)$ so $n\overline X=\sum_{i=1}^n X_i\sim\Gamma(n,1)$ and consequently $2n\overline X\sim\Gamma(n,\frac12)$.
The corresponding PDF is:$$\frac1{2^n\Gamma(n)}x^{n-1}e^{-\frac12x}$$ and is exactly the PDF of $\chi^2_{(2n)}$.
A: You can eliminate choices A and C immediately, since A does not depend on $n$, when it clearly should; and C is obviously wrong when $n = 1$, in which case $2n \bar X = 2X_1$, hence exponentially distributed, which $\chi^2_2$ is, but $\chi^2_1$ is not.  Finally, we can eliminate B because we know that the sample total is not exponentially distributed for $n > 1$.
Another way to look at it is to note that $\operatorname{E}[2n \bar X] = 2n$, thus the correct answer choice must also have the same mean.  If you have memorized the mean of the chi-square distribution, you'd know this eliminates C; but if not, you can reason that since the chi-square distribution with $n$ degrees of freedom is the sum of $n$ squared standard normal random variables, that its mean must be by linearity of expectation $$\operatorname{E}[Z_1^2 + \cdots + Z_n^2] \overset{\text{ind}}{=} n \operatorname{E}[Z_1^2] = n(\operatorname{Var}[Z_1] + \operatorname{E}[Z_1]^2) = n(1 + 0) = n.$$  But I prefer the first argument.
