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I would like to show the following:

$$\lim_{y \to +\infty} \left[\frac{1}{\sqrt{\pi}e^{y^2}(1-\text{erf}(y))}-y\right]=0.$$

Here $$\text{erf}(y)= \frac{2}{\sqrt{\pi}} \int_{0}^{y}e^{-x^2}dx$$ denotes the error-function. In fact I would like to have that $\lim\limits_{y \to +\infty} [\mathbb{E}(X\mid X \geq y)- y]=0$ provided that $X \sim \mathcal{N} (0,\frac{1}{\sqrt{2}})$. Since $$\mathbb{E}(X\mid X \geq y)= \frac{\int_{y}^\infty xe^{-x^2} dx}{\int_{y}^\infty e^{-x^2} dx}=\frac{\frac{1}{2}e^{-y^2}}{\frac{2}{\sqrt{\pi}}(1-\text{erf}(y))}=\frac{1}{\sqrt{\pi}e^{y^2}(1-\text{erf}(y))}$$ the second limit should then reduce to the first one in the yellow box.

Since $erf(y)$ cannot be stated explicitely with elementary functions I tried to use approximizations of it but it seems that it is not enough to allow errors up to a constant (although the constant is tiny) to show that the limit is zero. I tried the ones given here: https://en.wikipedia.org./wiki/Error_function#Approximation_with_elementary_functions.

I think that intuitively the statments should be correct because of the strong descent of the Gaussian function $e^{-x^2}$ when approaching infinity: The bigger $y$ is the more I expect $\mathbb{E}(X\mid X \geq y)$ to be close to $y$.

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  • $\begingroup$ By L'Hospital's rule, as $y \to \infty$, $$\frac{\frac{e^{-y^2}}{\sqrt{\pi}}}{(1-\text{erf}(y))}=y+\varepsilon(y)$$ with $\lim_{y \to \infty}\varepsilon(y)=0$, see my answer below. $\endgroup$ – Olivier Oloa Feb 15 '17 at 15:44
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Hint. One may observe that, as $y \to \infty$, $$ \frac{1}{\sqrt{\pi}e^{y^2}(1-\text{erf}(y))} =\frac{\frac{e^{-y^2}}{\sqrt{\pi}}}{(1-\text{erf}(y))}\to \frac00 $$ then one is allowed to apply L'Hôpital's rule getting, as $y \to \infty$, $$ \frac{\frac{e^{-y^2}}{\sqrt{\pi}}}{(1-\text{erf}(y))}\sim\frac{\frac{-2ye^{-y^2}}{\sqrt{\pi}}}{-\frac{2}{\sqrt{\pi}}e^{-y^2}}=y. $$

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