Question: Let $Z_{i}$ ~ $N(0,1)$ be independent random variables for i = 1,2,... . Obtain the limiting distribution of $$\frac{1}{\sqrt{n}}\Bigl(\sum_{i=1}^n\Bigl(Z_{i} + \frac{1}{n}\Bigr)\Bigr)$$ using moment generating functions.

So far I have: Let $Y_{n}=\frac{1}{\sqrt{n}}(\sum_{i=1}^n(Z_{i} + \frac{1}{n}))$

$\\M_{y_n}(t)=E(e^{ty_n})=E(exp(t[\frac{1}{\sqrt{n}}\sum_{i=1}^n(Z_{i} + \frac{1}{n})]))=...=e^\frac{t}{\sqrt{n}}\prod_{i=1}^nE(exp\frac{t}{\sqrt{n}}Z_{i})=e^\frac{t}{\sqrt{n}}(M_{Z_i}(\frac{t}{\sqrt{n}}))^n$

I'm not too sure where to go from here. I think I can apply Taylor expansion to $M_{Z_i}(\frac{t}{\sqrt{n}})$ at 0, but how do I work with the $\frac{t}{\sqrt{n}}$? Am I to take the derivative of that function and calculate it at zero? Likely a simple concept that I am not understanding, so any clarification would help, thanks!

  • $\begingroup$ I think $M_{Z_i}\left(\frac{t}{\sqrt{n}}\right)$ is just the moment generating function of $\frac{1}{\sqrt{n}} Z_i$, which is a normal with mean zero and variance $\frac{1}{n}$, no? If so, then you could look that up in a table, or use the Taylor expansion. $\endgroup$
    – Kevin
    Jul 4, 2017 at 22:48
  • $\begingroup$ Are you allowed to use the obvious reduction of the sum to a Normal$(1,n)$ variable or the equally obvious reduction of the whole thing to a Normal$(1/\sqrt{n},1)$ variable, or do you also have to work out those results using mgfs? $\endgroup$
    – whuber
    Jul 5, 2017 at 18:45

1 Answer 1


By the properties of Gaussian RVs (unrelated to MGFs),

$$ Z_i \sim \mathcal{N}(0, 1) \Rightarrow (Z_i + \frac{1}{n}) \sim \mathcal{N}\left(\frac{1}{n}, 1\right). $$

The MGF of this distribution is

$$ e^{\frac{t}{n} + t^2} $$

Since the RVs are independent, the weighted sum has a simple form

$$ \left( e^{\frac{t}{n\sqrt{n}} + \left(\frac{t}{\sqrt{n}}\right)^2} \right)^n = e^{\frac{t}{\sqrt{n}} + t^2} . $$

At the limit, this is $$ e^{t^2}, $$

which is the MGF of $\mathcal{N}(0, 1)$.

  • $\begingroup$ +1 -- but what is "$\sigma^2$"? $\endgroup$
    – whuber
    Jul 5, 2017 at 16:28
  • 1
    $\begingroup$ Should there not be a $\frac{1}{2}$ in front of the $t^{2}$ $\endgroup$
    – Quality
    Jan 23, 2019 at 4:44

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