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?


  • $\begingroup$ What on earth? Why is $Y_n$ suddenly a sum of $n$ independent normal random variables instead of a (displaced and scaled) sum of $n$ independent $\chi^2_1$ random variables? Are you applying the CLT in some vague hand-waving sense? $X_n \sim \chi^2_n$ has mean $n$ and variance $2n$ and so $Y_n = \frac{X_n-n}{2n}$ is a zero-mean random variable of variance $\frac{1}{4n^2}\times 2n = \frac{2}{n}$. Both your calculation of $E[Y_n]$ and var$(Y_n)$ are incorrect, and the limiting distribution of $Y_n$ is incorrect too: $Y_n$ converges to $0$, not $-\frac 12$. $\endgroup$ – Dilip Sarwate Apr 14 '15 at 15:00
  • $\begingroup$ Hi @DilipSarwate. I just realized I've completely misread the definition of a Chi squared distribution. I thought it was the sum of n standard normal distributions, rather than the sum of their squares! On retrospect I should of noticed that's not the case because the sum of normally distributed random variables is normally distributed. Thanks for your comment though, I will redo this problem with the proper definition and use your comment if I need help $\endgroup$ – HBeel Apr 14 '15 at 15:58

$X_n$ 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 $n$.

  • $\begingroup$ Thanks for your answer. I made a mistake with my definition of the Chi squared distribution which is my answer makes practically no sense. I'm aware of these properties, and I was using them in my answer (just in dragged out steps and a wrong understanding of $X_n$!). The trick for expectation of $\chi_1^2$ is useful to know as well, and I'm sure I can work out how to derive variance with a little bit of googling. Thanks @Glen_b $\endgroup$ – HBeel Apr 15 '15 at 9:48
  • $\begingroup$ With respect to the last sentence of your answer, you might want to consider adding just a little with regard to how the variance of a $\chi_1^2$ random variable is determined. $E[X_1]=E[Z^2]=1$ is easy enough since $Z\sim N(0,1)$, but to get $E[X_1^2]=E[Z^4]$ needs a little more work, or more knowledge of the moments of the standard normal random variable (or Gamma random variables since $\chi_1^2\sim \Gamma(\frac 12, \frac 12)$. $\endgroup$ – Dilip Sarwate Apr 15 '15 at 18:18

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