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Given $T_n$ is a discrete r.v. with pmf $\ f_{X_n}(t) = \frac{1}{n}$ for $x = n^2$, $ 1-\frac{1}{n}$ for $x=0$, and $\ 0$ otherwise. Define $T_n = X_n - E(X_n)$. Find the pmf (or pdf) of the limiting distribution.

My attempt: First, since $X_n$ is a discrete r.v, $E(X_n) = n^2\frac{1}{n} + 0(1-\frac{1}{n}) + 0 = n$. Thus $T_n = X_n - n$. Now, $F_{T_n}(t) = P(T_n\leq t) = P(X_n\leq n+t) = 1$ if $t\geq n^2-n$, $= 1-\frac{1}{n}$ if $-n\leq t < n^2 - n$ and $= 0$ if $t< -n$.

Thus, as $n\rightarrow \infty$, $F_{T_n}(t) = 1$ for $t\in (-\infty, \infty)$ (is this a correct interval?), we conclude that $T_1, T_2, \ldots$ converges in distribution to a degenerate r.v $T$ whose pmf is $f_{T}(t) = 1$.

My question: Could someone please help verify if my solution above is correct?

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    $\begingroup$ What precisely is given: pmf $f_{T_n}(t)$ or $f_{X_n}(t)$? $\endgroup$
    – NCh
    Mar 2, 2017 at 2:00
  • $\begingroup$ I messed up a bit. Thank you for pointing that out. $\endgroup$
    – ghjk
    Mar 2, 2017 at 2:47
  • $\begingroup$ @NCh: please help give this problem a try, as I'm still pondering a bit about the conclusion made by Viktor in the comment below. $\endgroup$
    – ghjk
    Mar 2, 2017 at 5:12
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    $\begingroup$ @ Oh, yes. And how can you interpret your result: $F_{T_n}(t)\to 1$ for all $t\in\mathbb R$? Does the proper limiting distribution exists? Can $F_T(t)=1$ for all $t$ be a CDF of a r.v. taking values in $\mathbb R$? $\endgroup$
    – NCh
    Mar 2, 2017 at 14:55
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    $\begingroup$ I'm..so stupid:( Thanks a ton. The problem is when $t\rightarrow -\infty$. $\endgroup$
    – ghjk
    Mar 2, 2017 at 16:34

1 Answer 1

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The limiting distribution is $-\infty$ with probability 1. That is what $F(t) = 1$ for $t\in (-\infty, \infty)$ means.

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  • $\begingroup$ thanks for your help, but how did you get that conclusion?? I meant, a $-\infty$ limiting distribution is not a valid pdf, isn't it? $\endgroup$
    – ghjk
    Mar 1, 2017 at 18:48
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    $\begingroup$ It is not a proper distribution, but it is a valid degenerate distribution. $\endgroup$
    – Viktor
    Mar 1, 2017 at 18:50
  • $\begingroup$ Thanks for your prompt explanation. So the range of $t$ is still $(-\infty, \infty)$ in this case? How could we integrate that kind of function to get $F(t) = 1$ then? $\endgroup$
    – ghjk
    Mar 1, 2017 at 18:51
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    $\begingroup$ The limit pmf is 1 at $-\infty$ and zero elsewhere. $\endgroup$
    – Viktor
    Mar 1, 2017 at 18:53
  • $\begingroup$ Hmm...this is quite interesting result. Thanks for your patience. Btw, could you please help check my conclusion in the comment for this problem as well: math.stackexchange.com/questions/2166528/…? $\endgroup$
    – ghjk
    Mar 1, 2017 at 20:13

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