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The following question I found on an old exam: Given $n$ i.i.d. random variables $X_k$, $1 \leq k \leq n$, with uniform distribution on $[-1,1]$, it is easy to compute the characteristic function of one them, namely $$ \phi_{X_k}(t) = \frac{\sin (t)}{t}.$$ By the properties of the characteristic function and by independence, we also have, $$\phi_{\bar{X_n}}(t) = \left[\frac{\sin(\frac{t}{n})}{\frac{t}{n}}\right]^n.$$ Then finally one is asked, to show that $$ \lim_{n \rightarrow \infty}\left[\frac{\sin(\frac{t}{\sqrt{n}})}{\frac{t}{\sqrt{n}}}\right]^n \rightarrow \exp\left(-\frac{t^2}{6}\right),$$ by using the Central Limit theorem. We are also given $\mathbb{E}(X_k^2) = \frac{1}{3}$ and the hint that if $Y \sim \mathcal{N}(0, \sigma^2)$, then $\phi_Y(t) = \exp\left(-\frac{(\sigma t)^2}{2}\right).$

This is where I've been struggling so far. Any input would be greatly appreciated :) Thanks!

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    $\begingroup$ Are you supposed to show this directly, or by showing that the CLT applies to $U(-1,1)$ variates, using this to show that the limit of the ch.f. of the sum is $\exp(-3t^2/2)$, and deducing from this that the limit on the left is the right? $\endgroup$
    – jbowman
    Nov 20, 2017 at 18:16
  • $\begingroup$ The latter would be sufficient... $\endgroup$ Nov 20, 2017 at 18:18
  • $\begingroup$ I'd be much obliged if you could help in any way :) $\endgroup$ Nov 20, 2017 at 18:21
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    $\begingroup$ As always when taking a limit of the form $\lim_{n\to\infty} f(tn^\alpha)^n$, take logarithms and use the first couple of terms in a power series in $n^\alpha$. $\endgroup$
    – whuber
    Nov 20, 2017 at 18:38
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    $\begingroup$ Given that you are able to use the CLT, it seems to me you could just derive the asymptotic mean and variance, use the CLT to state what the limit ch.f. is given that, and then you have the r.h.s. of the expression. Depending on how the question is worded, that might be sufficient - but it's better practice to work through the limits as @whuber suggests. $\endgroup$
    – jbowman
    Nov 20, 2017 at 19:36

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There are several ways you could do this, but one is to expand the sine function using its Maclaurin expansion, which gives:

$$\text{sinc}(x) = \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots.$$

This gives you:

$$\text{sinc}(\tfrac{t}{\sqrt{n}}) = 1 - \frac{t^2/6}{n} + \frac{t^4/120}{n^2} - \cdots.$$

Since the higher-order terms vanish in the limit you have:

$$\lim_{n \rightarrow \infty} \Big[ \text{sinc}(\tfrac{t}{\sqrt{n}}) \Big]^n = \lim_{n \rightarrow \infty} \Big[ 1 - \frac{t^2/6}{n} \Big]^n = \exp \Big( -\frac{t^2}{6} \Big),$$

where we use Bernoulli's limiting definition of $e$ in the last step.

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  • $\begingroup$ How come $t^2/(6n)$ does not vanish? $\endgroup$ Nov 27, 2018 at 10:54
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    $\begingroup$ It can be shown that as $n \rightarrow \infty$ we get $(1+\tfrac{x}{n})^n \rightarrow e^{x}$, so the result follows with $x = -t^2/6$. If you would like to know more about this, I would recommend having a read about this limiting definition of $e$. $\endgroup$
    – Ben
    Nov 27, 2018 at 10:57
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    $\begingroup$ you're right, that's quite standard... $\endgroup$ Nov 27, 2018 at 10:58

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