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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.

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    $\begingroup$ 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? $\endgroup$
    – P.Windridge
    Apr 23, 2015 at 16:50
  • $\begingroup$ @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. $\endgroup$
    – Nicky_Ay
    Apr 24, 2015 at 3:54
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    $\begingroup$ 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. $\endgroup$
    – Alecos Papadopoulos
    Apr 24, 2015 at 3:58
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    $\begingroup$ 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. $\endgroup$
    – Christoph Hanck
    Apr 24, 2015 at 4:43
  • $\begingroup$ possible duplicate of Is there a theorem that says that $\sqrt{n}\frac{\bar{X} - \mu}{S}$ converges in distribution to a normal as $n$ goes to infinity? $\endgroup$
    – Alecos Papadopoulos
    Apr 24, 2015 at 9:51

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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.

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  • $\begingroup$ Thank you so much! This was the most clear answer I've gotten on CrossValidated so far $\endgroup$
    – Nicky_Ay
    Apr 24, 2015 at 14:52
  • $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Apr 24, 2015 at 15:27
  • $\begingroup$ 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. $\endgroup$
    – mpiktas
    Apr 24, 2015 at 18:09
  • $\begingroup$ 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. $\endgroup$
    – mpiktas
    Apr 24, 2015 at 18:13

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