Is this the correct way to calculate the mean and variance of $(X-n)/(2n)$ where $X$ follows a Chi-squared distribution with $n$ degrees of freedom? Let $X_n$ follow a Chi-squared distribution with $n$ degrees of freedom. I would like to study the variance and mean of $$\lim\limits_{n \rightarrow \infty} Y_n := \frac{(X_n-n)}{(2n)}$$
I would also like to characterise the behaviour of $Y_n$ for large degrees of freedom
To tackle this problem I observe $$X_n = \sum\limits_{k=1}^n N_i \quad \text{Where} \quad N_i \sim \mathcal{N}(0,1)$$
Hence 
\begin{align}
Y_n &= \frac{ \left( \sum\limits_{k=1}^n N_i \right) - n}{2n} \\
Y_n &= \frac{ \sum\limits_{k=1}^n N_i }{2n} - \frac{1}{2}
\end{align}
Then using linearity of expectation
\begin{align}
\mathbb{E}(Y_n) &= \mathbb{E} \left( \frac{ \sum\limits_{k=1}^n N_i }{2n} - \frac{1}{2} \right) \\
&= \frac{ \mathbb{E} \left(  \sum\limits_{k=1}^n N_i \right)  }{2n} - \frac{1}{2}  \\
&= -\frac{1}{2}
\end{align} 
Since $\forall n \in \mathbb{N} $ we have $\mathbb{E}(Y_n) = - \frac{1}{2}$ then $\lim\limits_{n \rightarrow \infty} \mathbb{E}(Y_n) = - \frac{1}{2}$
Now using $\mathbb{V}(aX + b) = a^2 \mathbb{V}(X)$ and linearity of variance for independent variables I find
\begin{align}
\mathbb{V}(Y_n) &= \mathbb{V} \left( \frac{ \sum\limits_{k=1}^n N_i }{2n} - \frac{1}{2} \right) \\
&= \mathbb{V} \left( \frac{ \sum\limits_{k=1}^n N_i }{2n} \right) \\
&= \frac{1}{4n^2} \mathbb{V}  \left(  \sum\limits_{k=1}^n N_i  \right) \\
&= \frac{1}{4n^2}    \sum\limits_{k=1}^n \mathbb{V} (N_i)   \\
&= \frac{n}{4n^2} = \frac{1}{4n}
\end{align}
So $\lim\limits_{n \rightarrow \infty} \mathbb{V}(Y_n) = \lim\limits_{n \rightarrow \infty} \frac{1}{4n} =0$.
My question is have I calculated the mean and variance of $Y_n$ correctly? 
And also as a side question does $Y = \lim\limits_{n \rightarrow \infty} Y_n$ define a distribution?
Thanks.
 A: $X_n$ is a sum of $n$ $\chi^2_1$ random variates, or the sum of the squares of $n$ standard normals. 
With the expectation, if you don't claim that each component in $X_n$ (i.e. $N_i$) is normal (it isn't) or that it has mean 0 and variance 1 (it doesn't), you could be correct up to here:
$\begin{align}
\mathbb{E}(Y_n) &= \mathbb{E} \left( \frac{ \sum\limits_{k=1}^n N_i }{2n} - \frac{1}{2} \right) \\
&= \frac{ \mathbb{E} \left(  \sum\limits_{k=1}^n N_i \right)  }{2n} - \frac{1}{2}  
\end{align}$
But the next step after that is not correct (because that's where you use the things you had wrong). You seem to be suggesting that the expectation in the above line is 0. 
Instead (if you need to do it this way at all), let $N_i$ be $\chi^2_1$ and let $N_i=Z_i^2$ where the $Z_i$ are independent standard normal r.v.s.  Then $X_n=\sum N_i=\sum Z_i^2$.
Note that the expectation of a chi-square with 1 df is 1 ... $E(Z_i^2)=Var(Z_i)+E(Z_i)^2=1+0=1$.
It looks like you're making your life hard. You just apply some simple properties of expectation and variance.
I'll take as known that a $\chi^2_\nu$ random variable that it has mean $\nu$ and variance $2\nu$.*
$E(\frac{X_n-n}{2n})=\frac{1}{2n}[E(X_n)-n]$ ... but recall we know $E(X_n)$ already.
$\text{Var}(\frac{X_n-n}{2n})=\frac{1}{4n^2}\text{Var}(X_n)$ ... and again, use the known variance.
In each case we're simply applying a couple of the properties I linked to (which I assume you already were aware of).
* If we don't take it as given, you can show it for a $\chi^2_1$ and then apply the above simple rules again to establish it for $n$.
