# Limiting distribution of $T_n=\frac{\sum_{i=1}^n(2X_i−n)}{n}−3$ where $f_k(x) =\frac{x^k}{k!}e^{−x}I_{(0,\infty)}(x)$

$$X_1$$, $$X_2$$,..., are independent random variables. $$X_k$$ has pdf

$$f_k(x) =\frac{x^k}{k!}e^{−x}I_{(0,\infty)}(x)$$

Let

$$T_n=\frac{\sum_{i=1}^n(2X_i−n)}{n}−3$$

Consider the sequence $$T_1$$, $$T_2$$,.., and give the pmf or pdf of the limiting distribution.

I tried to use the following theorem:

$$\lim_{n\rightarrow\infty}M_{T_n}(t)=M_T(t)$$ for all $$t$$ in a neighborhood of $$0$$ $$\Rightarrow T_n \rightarrow T$$

I first note that $$f_k(x)\sim Gamma(\alpha=k+1,\beta=1)$$ which has mgf $$M_{X_k}(t)=\left(\frac{1}{1-t}\right)^{k+1}$$

Then

\begin{align*} M_{T_n}(t) &=M_{\frac{\sum(2X_i−n)}{n}−3}(t)\\\\ &=M_{\frac{2\left(\sum X_i\right)−n^2}{n}−3}(t)\\\\ &=M_{\frac{2}{n}\left(\sum X_i\right)-n−3}(t)\\\\ &=e^{-(n+3)t}M_{\sum X_i}\left(\frac{2t}{n}\right)\\\\ &=e^{-(n+3)t}M_{X_1}\left(\frac{2t}{n}\right)\cdot M_{X_2}\left(\frac{2t}{n}\right)\cdots M_{X_n}\left(\frac{2t}{n}\right)\\\\ &=e^{-(n+3)t}\cdot\left(\frac{1}{1-\frac{2t}{n}}\right)^{2}\cdot \left(\frac{1}{1-\frac{2t}{n}}\right)^{3}\cdots \left(\frac{1}{1-\frac{2t}{n}}\right)^{n+1}\\\\ &=e^{-(n+3)t}\cdot\left(\frac{1}{1-\frac{2t}{n}}\right)^{\frac{(n+1)\cdot(n+2)}{2}-1} \end{align*}

Then the next step would be to take

$$\lim_{n\rightarrow\infty}\left(e^{-(n+3)t}\cdot\left(\frac{1}{1-\frac{2t}{n}}\right)^{\frac{(n+1)\cdot(n+2)}{2}-1}\right)$$

which looks rather messy. I am afraid I made a mistake somewhere. Is there a more simple approach to finding the limiting distribution?

I have also tried taking

\begin{align*} F_{T_n}(t) &=P(T_n\leq t)\\\\ &=P\left(\frac{\sum(2X_i−n)}{n}−3\leq t\right)\\\\ &=P\left(\sum X_i \leq \frac{n(t+3)+n^2}{2}\right)\\\\ \end{align*}

but I don't think this method wouldn't work as the $$X_k$$'s are not identically distributed.

I'm looking for a nudge in the right direction. Hopefully then I can finish it from there.

• Are you saying that $(k+1)E\sim X_k$? – Remy Sep 23 '18 at 21:36
• Sorry, my bad, it should be $\sum_1^{k+1} E_i$ – InfProbSciX Sep 23 '18 at 21:39
• Yeah, sorry about that, my answer above is wrong, you'll have to do a double sum instead: $$X_i = \sum_i^{k+1} E_i$$ ... where E_i is an exponential distribution, here's the reference: "If X ~ Gamma(α, θ) and Y ~ Gamma(β, θ) are independently distributed, then ... X + Y, which is Gamma(α + β, θ)-distributed" So: $$\sum_i X_i = \sum_i \sum_1^{i+1} E_i$$ – InfProbSciX Sep 23 '18 at 21:43
• So then would $E_{i}\sim exp(i)$? If so, wouldn't that lead to the same problem as before? – Remy Sep 23 '18 at 21:44
• Not necessarily, you will be able to write that sum as one sum (instead of a double sum) of independent exponentials, which will be a single gamma distribution. Btw, $E_i \sim Exponential(1)$, not i. – InfProbSciX Sep 23 '18 at 21:47