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Given that the characteristic function for Y is $$ \varphi_Y (t) = e^{\lambda (e^{-t^2/2}-1)} $$

Show that

$$ Y/\sqrt{\lambda } \xrightarrow{d} N(0,1) $$ as $$ \lambda \rightarrow \infty $$

I've tried to find the characteristic function of $Y/\sqrt\lambda$ to show convergence from there, but I am getting stuck. Here is what I have tried:

$$ \varphi_{Y/\sqrt\lambda}(t) = E[e^{ty/\lambda}] = e^{\lambda (e^{-t^2/2}-1)}\bullet\varphi_Y(t/\sqrt\lambda) \\=e^{\lambda (e^{-t^2/2}+e^{-t^2/2\lambda}-2)} $$

But when I try and take the limit of this function it is way to complicated and I keep hitting a dead end. Any hints or ideas would be helpful!

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  • $\begingroup$ Thanks saz. This was very helpful, but I'm having difficulty following the part where algebraically manipulate the exponential. How are you evaluating the derivative at zero to solve the limit? I admit my calculus is rusty. $\endgroup$ – Mark Jun 26 '15 at 18:54
  • $\begingroup$ By the very definition of the derivative, we have $$\frac{\exp(-h)-1}{h} \to \frac{d}{dx} \exp(-x) \bigg|_{x=0} = -1$$ as $h \to 0$. Use this identity for $h := \frac{t^2}{2\lambda}$. $\endgroup$ – saz Jun 26 '15 at 19:52
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There is a mistake in the computation of the characteristic function. By definition,

$$\varphi_Z(t) = \mathbb{E}e^{\imath \, t Z}, \qquad t \in \mathbb{R}.$$

Therefore,

$$\begin{align*} \varphi_{\frac{Y}{\sqrt{\lambda}}}(t) &= \mathbb{E} \exp \left( \imath \frac{t}{\sqrt{\lambda}} Y \right) = \exp \left( \lambda \left[ \exp \left(- \frac{t^2}{2\lambda} \right)-1 \right] \right) \end{align*}$$

where we have used in the last step that the characteristic function of $Y$ is given. Since

$$\lambda \left[ \exp \left(- \frac{t^2}{2\lambda} \right)-1 \right] = \frac{t^2}{2} \frac{\exp \left(- \frac{t^2}{2\lambda} \right)-1}{\frac{t^2}{2\lambda}},$$

we find

$$\lim_{\lambda \to \infty}\lambda \left[ \exp \left(- \frac{t^2}{2\lambda} \right)-1 \right] = \frac{t^2}{2} \frac{d}{dx} \exp \left(-x \right) \bigg|_{x=0} = - \frac{t^2}{2}.$$

This proves

$$\lim_{\lambda \to \infty}\varphi_{\frac{Y}{\sqrt{\lambda}}}(t) = \exp \left( - \frac{t^2}{2} \right) \qquad \text{for all $t \in \mathbb{R}$}.$$

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