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I want to find the limits in $\infty, $ and $-\infty$ of the function $f(z)= (z\Phi + \phi )(2\Phi^2-2\Phi+1)-z\Phi^2$ where $\phi=\phi(z)$ is the pdf of a standard normal variable, and $\Phi=\Phi(z)$ is the cdf of a standard normal variable. I know this limits are zero in both cases, but I want to prove it formally.

This is what I have done, for the case in which $z \rightarrow \infty$,

$\lim_{z\rightarrow \infty} f(z)= \lim_{z\rightarrow \infty} 2z\Phi^3-2z\Phi^2+z\Phi+2\phi\Phi^2-2\phi \Phi+\phi-z\Phi^2$

which is equal to $\lim_{z\rightarrow \infty} 2z\Phi^3-3z\Phi^2+z\Phi + \lim_{z\rightarrow \infty} 2\phi \Phi^2-2\phi \Phi+\phi=\lim_{z\rightarrow \infty}2z\Phi^3-3z\Phi^2+z\Phi+0$

This limit

$\lim_{z\rightarrow \infty} 2z\Phi^3-3z\Phi^2+z\Phi=\lim_{z\rightarrow \infty} z (2\Phi^3-3\Phi^2+\Phi)$ is of the form $\infty \times 0$ so I have tried to write the expression as:

$\lim_{z\rightarrow \infty} z/({\frac{1}{(2\Phi^3-3\Phi^2+\Phi)}})$ and use Hopital rule, but I am getting things of the form $\frac{0}{0}$ since the derivatives of $\Phi$ have the term $\phi$ and this is zero in the limit.

Thanks in advance

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1 Answer 1

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You can use an easy estimate, with

$$\phi(t) = \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}} \qquad \Phi(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z e^{-\frac{t^2}{2}} d t$$

When $t \ge 1$ one has $e^{-\frac{t^2}{2}}\le t e^{-\frac{t^2}{2}}$, hence $$ z\ge 1 \Longrightarrow \frac{1}{\sqrt{2\pi}}\int_z^{+\infty} e^{-\frac{t^2}{2}} d t \le \frac{1}{\sqrt{2\pi}}\int_z^{+\infty} t e^{-\frac{t^2}{2}} d t = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} = \phi(z) $$ It follows that $$z\ge 1 \Longrightarrow 0\le (1-\Phi(z))\le \phi(z)$$ Now since $\lim_\limits{z\to+\infty}z^k\phi(z) = 0$ for any $k\in {\mathbb R}$, we have $\lim_\limits{z\to+\infty}z^k (1-\Phi(z)) = 0$

When $z\to - \infty$, we know that $\Phi(z)= (1 - \Phi(-z))$, hence $\lim_\limits{z\to-\infty}|z|^k \Phi(z) = 0$.

Now observe that $2\Phi^3 - 3 \Phi^2+ \Phi = \Phi(1 -\Phi)(1 - 2\Phi)$. It follows easily that $|z|^k (2\Phi^3 - 3 \Phi^2+ \Phi)\to 0$ when $z\to \pm\infty$.

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