1

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
  • 851
  • 8
  • 18
  • 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
  • 2
    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
  • 1
    "For large $n$" and "approximate" should both bring to mind the Central Limit Theorem. One way to handle the square root would be with the Delta method, 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
  • 1
    @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

0 Answers0