# sum of random variables

I have a random variable $T$ defined as $T=\frac{1}{n}(X_1+\cdots+X_n)$ with $X_i$ being $N(\mu,1)$ distributed. I want to know how $T$ is distributed. This is what I did but I am not sure whether its good: I know that the sum of $n$ normal distributed random variables is distributed $N(n\mu,n\sigma^2)$. So $X_1+\cdots+X_n=n\mu+n\sigma U$ with $U$ being the standard normal distribution. So $T=\mu+U$.

Edit:I think something is wrong because $T$ is indepent of $n$. So what is wrong?

Is this the good way to solve such questions?

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$X_1,\ldots,X_n$ are independent, right? – Stefan Hansen Jan 20 '13 at 15:53
@StefanHansen yes they are independent. – Badshah Jan 20 '13 at 15:54

This is not entirely true. You have that $X_1+\cdots +X_n=n\mu +\sqrt{n}\sigma U$ for some random variable $U$ following a standard normal distribution, and so $T=\mu+\frac{\sigma}{\sqrt{n}}U$.
However, I like doing it this way where you don't have to introduce $U$: $$X_1+\cdots X_n\sim N(n\mu,n\sigma^2),$$ where $\sim$ means "is distributed" and so $$T=\frac{1}{n}\left(X_1+\cdots+X_n\right)\sim N\left(\mu,\frac{1}{n}\sigma^2\right).$$ This way you're only talking about the distribution of $T$ and not about any pointwise equality between $T$ and some other random variable $U$.
Another way to go about it is to note that, by independence, we have that $(X_1,\ldots,X_n)$ follows a multivarite normal distribution and so we know that any linear transformation of this vector is again a normal distribution. And since $T=f(X_1,\ldots,X_n)$, where $f$ is the linear function $$f(x_1,\ldots,x_n)=\frac{1}{n}(x_1+\cdots+x_n),$$ we already know that $T$ is normal distributed. So we just have to determine its mean and variance, and this can be done by properties of the expectation, i.e. $$E[T]=\frac{1}{n}\left(E[X_1]+\cdots +E[X_n]\right)=\frac{1}{n}nE[X_1]=\mu,$$ and by independence $$\mathrm{Var}(T)=\frac{1}{n^2}\left(\mathrm{Var}(X_1)+\cdots+\mathrm{Var}(X_n)\right)=\frac{1}{n^2}n\sigma^2=\frac{1}{n}\sigma^2.$$ Hence $T\sim N(\mu,\frac{1}{n}\sigma^2)$.