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 ,

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


  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Nov 2 '13 at 1:17
  • 2
    $\begingroup$ 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 $f(y=0)$. $\endgroup$ Nov 2 '13 at 1:47
  • $\begingroup$ Was the second term perhaps supposed to be $2a(1-a)$? Or was maybe the last term just supposed to be $1-a$? $\endgroup$
    – Glen_b
    Nov 2 '13 at 1:52
  • 1
    $\begingroup$ "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.) $\endgroup$
    – whuber
    Nov 2 '13 at 13:13
  • 1
    $\begingroup$ @whuber Shouldn't $\mu_n$ be $2(1-a)$ $\endgroup$
    – user30438
    Nov 2 '13 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.