# Slutsky's Theorem to show convergence to Standard Normal Distribution

We are given $W_n = \frac{\bar{X}-\lambda}{\sqrt{\bar{X}/{n}}}$ and need to show it converges to a standard normal distribution.

EDIT: The square root in my original post did not extended over the $n$ in the denominator as well. It has been fixed.

I want to use Slutky's theorem which says that if we have two sequences $X_n$ and $Y_n$ which converge respectively to some random variable $X$ and come constant $c$ then $X_n \times Y_n$ will converge to $Xc$.

With this in mind, I multiply my $W_n$ by $\frac{\sqrt{\bar{X}}}{\sqrt{n\sigma}}$

EDIT: I would multiply by $\frac{\sqrt{\bar{X}}}{\sigma}$ instead.

This would mean that my sequence $X_n = \frac{\bar{X} - \lambda}{\sigma/\sqrt{n}}$ would have the form which we know converges to $N(0,1)$ by Central Limit Theorem.

And it would remain to show what $Y_n = \frac{\sqrt{\bar{X}}}{\sigma}$ converges to.

I'm not sure how to handle that last step.

Also, if there is a better way to approach the problem I would appreciate feedback. Thank you.

• Hmm, assuming $\bar X$ is the mean, then it can be negative. So what is $\sqrt{\bar X}$? Do you rather want the sample variance? Apr 23, 2015 at 16:50
• @P.Windridge What I wrote in the first line was given to me in the question. I agree with what you are saying but what I typed is exactly how the question was phrased. Apr 24, 2015 at 3:54
• If $\lambda$ is the common mean of the $X$'s, then we can take $X$ to be a positive random variable, in which case its square root is defined in the reals. Apr 24, 2015 at 3:58
• a guess: do we talk about a Poisson problem - $\lambda$ is common notation for the parameter, and mean=variance for this distribution. Also: the very first square root in the first line should probably stretch over $n$, too. Apr 24, 2015 at 4:43
• Apr 24, 2015 at 9:51

For $Y_n$ you should use law of large numbers and continuous mapping theorem, i.e. if $Z_n\to Z$ in probability, then $g(Z_n)\to g(Z)$ in probability for continuous $g$.
You have $Y_n=\frac{\sqrt{\bar X}}{\sqrt{n}\sigma}$. Due to LLN $\bar X\to\lambda$, so the nominator converges to $\sqrt{\lambda}$. The denominator however converges to the infinity, hence the limit of the fraction is zero. However if denominator of $W_n$ is $\sqrt{\frac{\bar X}{n}}$ instead of $\frac{\sqrt{\bar X}}{n}$, the $Y_n=\frac{\sqrt{\bar X}}{\sigma}$ and the end result is $\frac{\sqrt{\lambda}}{\sigma}$, which is more feasible than zero.
• It's pretty murky to me, because I have the same difficulty expressed by @P.Windridge in a comment: several additional assumptions are needed, not least of which is that the $X_n$ be non-negative variables.
• If $X_i$ are positive and iid, $\bar X=\sum_{i=1}^n X_i$ and $\lambda = EX_i$, then everything checks out. This seems like a textbook exercise for students to test their understanding of CLT, LLN and Slutsky lema. The OP simply lacks necessary mathematical rigour to make the question plausible. Apr 24, 2015 at 18:09
• Oh and I noticed that my answer is incomplete. I only answered the part what to do with $Y_n$. As it stands now the limit is normal distribution with zero mean, but not unit standard deviation. I' ll fix the answer when I get the chance to use the computer, Android keyboard is not really suitable for Latex. Apr 24, 2015 at 18:13