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Given $ Y_1, Y_2..Y_n$ are iid from a distribution with pmf,
$f(y) = a^{2}$ for $y=0$,

$f(y) = 2a(1-a)$ for $y=1$ ,

$f(y) = (1-a)^{2}$ for $y=2$, where $0<a<1$.

For large n, calculate the approximate distribution of

a) $\sqrt {\bar{Y}}$ - Solution to part(a) posted as answer(awaiting confirmation)

b) $\sqrt n ({\bar{Y}-\mu)}+\bar Y^2$ , where $\mu=E(Y_1)$

Could you please verify my solution for part (b) :

By CLT $\sqrt n ({\bar{Y}-\mu)} \rightarrow N(0,\sigma^2)$ (convergence in probability)

For $\bar Y^2$, applying delta method, $\bar Y^2 \rightarrow N(\mu^2,\frac{4\mu^2\sigma^2}{n^2})$ (converges in distribution)

{EDIT} - Can I say : $\bar Y^2 \rightarrow \mu^2$ in probability

where $\sigma^2 = Var Y$ and $Var \bar Y^2 = \sigma^2/n$

Can I apply slusky theorem, as one distribution converges in probability and other in distribution:

By Slutsky theorem ,

$\sqrt n ({\bar{Y}-\mu)} + \bar Y^2 \rightarrow [\mu^2 + N(\mu^2,\frac{4\mu^2\sigma^2}{n^2})]$

Thanks!

user30438
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  • If this is for study purposes, such as for some subject, please add the `self-study` tag. What is $X$? It hasn't been defined, only the $Y$'s have. – Glen_b Nov 02 '13 at 01:17
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    This cannot be a probability mass function. The total probability mass postulated is $a^2 + a(1-a) + (1-a)^2 = 1-a+a^2$. For this to equal $1$ we must have $a^2=a \Rightarrow a=1$, which concentrates all probability mass at $f(y=0)$. – Alecos Papadopoulos Nov 02 '13 at 01:47
  • Was the second term perhaps supposed to be $2a(1-a)$? Or was maybe the last term just supposed to be $1-a$? – Glen_b Nov 02 '13 at 01:52
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    "For large $n$" and "approximate" should both bring to mind the [Central Limit Theorem.](http://stats.stackexchange.com/questions/36113/where-does-sqrtn-come-from-in-central-limit-theorem-clt) One way to handle the square root would be with the [Delta method](http://stats.stackexchange.com/questions/5782/variance-of-a-function-of-one-random-variable), which in this case comes down to observing that with high probability $\sqrt{\bar{Y}} = \sqrt{\mu_n}\left(1 + (\bar{Y}-\mu_n)/(2\mu_n) + O(n^{-2})\right)$ where $\mu_n = n(1-a).$ (Applying Chebyshev's Inequality will make this rigorous.) – whuber Nov 02 '13 at 13:13
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    @whuber Shouldn't $\mu_n$ be $2(1-a)$ – user30438 Nov 02 '13 at 14:20
  • @user30438 You are correct: I typed an "$n$" instead of the $2$. The expectation of each $Y_i$ is, by definition, $0(a^2) + 1(2a(1-a))+2(1-a)^2=2(1-a),$ which therefore is the expectation of $\bar{Y},$ and that is what $\mu_n$ is supposed to be. Thanks for catching that! – whuber Nov 02 '13 at 14:25
  • @whuber I have posted my solution, could you please confirm whether it is correct. Does for approximation of distribution, I need to compute my solution till that step or are there any further steps involved. – user30438 Nov 02 '13 at 15:02
  • Please post your solution as an *answer*: this will bring it to the attention of everyone who visits here and provide a mechanism for them to comment, vote on, and even improve it. When you do that, please take care to recalculate the variance of $\bar{Y}$ and check your value with your intuition: as $n$ grows large, what ought to happen to the spread of that distribution? – whuber Nov 02 '13 at 15:03
  • @whuber I cross-checked my calculations, and did not find any error in variance. I calculated it using $Var(\sqrt{\bar Y})\approx [g'(\mu)]^2\sigma^2$ Could you please point where am I wrong. – user30438 Nov 02 '13 at 16:15

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