# How do I evaluate $k'$ from $\int_{k'}^{\infty} \sqrt{n/2\pi} \, \, e^{(-n/2 \, (\bar x-\mu_0)^2)} \, d\bar x = Z_\alpha$?

$k'$ is supposed to be $\mu_0+ \frac{Z_\alpha}{\sqrt n}$, but I don't know how to get there.

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I don't think the solution $k'=\mu_0+Z_\alpha/\sqrt n$ is correct. The solution involves error functions. Are you sure the integral is written correctly? (BTW, what have you tried?) – KennyTM Oct 2 '12 at 10:28
@KennyTM My dad was reading from one of his books and he asked me to find how to get to that value of $k'$ from the equation in there. This is a little ahead of what I'm used to figuring out and I've not tried anything useful so I decided to ask here instead. The value for $k'$ is correct because the author write "But we know $k'=...$" after he writes down the exact equation above. Could it be that it's not possible to extract that value of $k'$ from the integral itself hence the word 'but' and it's taken from whatever was going on before? – John Doe Oct 2 '12 at 10:50
Is there any chance that $Z_\alpha$ is some constant value like 0.35958045205206457? – KennyTM Oct 2 '12 at 10:57
@KennyTM $Z_\alpha = 1.645$ – John Doe Oct 2 '12 at 11:02
Can you give a decent title to this question? – Asaf Karagila Oct 2 '12 at 11:20

The change of variable $\bar x=\sqrt{n}(s-\mu_0)$ yields $$Z_\alpha=\int_u^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{-s^2/2}\mathrm ds,$$ with $u=\sqrt{n}\cdot(k'-\mu_0)$, hence $$2Z_\alpha=1+\mathrm{erf}(u/\sqrt2),$$ where $\mathrm{erf}$ denote the error function. In particular, $$k'=\mu_0+\sqrt{2/n}\cdot\mathrm{erf}^{-1}(2Z_\alpha-1).$$ Thus, $k'=\mu_0+Y_\alpha/\sqrt{n}$, as required, but for $Y_\alpha=\sqrt{2}\cdot\mathrm{erf}^{-1}(2Z_\alpha-1)$.

In particular, $Y_\alpha=Z_\alpha$ if and only if $Z_\alpha=z^*$ with $z^*\approx0.783264\ldots$, $Y_\alpha\lt Z_\alpha$ for every $Z_\alpha\lt z^*$ and $Y_\alpha\gt Z_\alpha$ for every $Z_\alpha\gt z^*$.

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Take a look here: Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$

You can use the same technique as Ross Millikan.

To be more specific:

Call $$I=\frac{n}{2\pi}\int_{k'}^{\infty}\int_{k'}^{\infty}e^{\frac{-n(x-\mu_{0})^{2}}{2}}dx$$

So $I^{2}=Z_{\alpha}^{2}$. But $$I^{2}=\frac{n}{2\pi}\int_{k'}^{\infty}\int_{k'}^{\infty}e^{-\frac{\bigl[\sqrt\frac{n}{2}(x-\mu_{0})\bigr]^{2}}{2}-\frac{\bigl[\sqrt\frac{n}{2}(y-\mu_{0})\bigr]^{2}}{2}}dxdy$$

Define $u=\sqrt\frac{n}{2}(x-\mu_{0})$ and $v=\sqrt\frac{n}{2}(y-\mu_{0})$, then

$$I^{2}=\frac{n^{2}}{4\pi}\int_{\sqrt\frac{n}{2}(k'-\mu_{0})}^{\infty}\int_{\sqrt\frac{n}{2}(k'-\mu_{0})}^{\infty}e^{-u^2-v^2}dudv$$

Now use polar coordinates and conclude.

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 Before I do, have you checked to see if you do indeed get the required value for $k'$? – John Doe Oct 2 '12 at 11:56 Sorry, i forgot to change the index... – Tomás Oct 2 '12 at 12:22 Sorry my answer for this question is not correct. I had did some mistakes here. I cant continue now from the last point... – Tomás Oct 2 '12 at 13:21