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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 $t< -n$.

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

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