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I'm looking to prove that $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$ using the fact that $E(Z^2)=\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}} z^2\, dz$ (where $Z$ is a standard normal variable), using the fact that $\Gamma(r)=\int_{0}^{\infty}y^{r-1}e^{-y}\,dy$.

The way I've gone about this is to allow $y={\frac{z^2}{2}}$ and so $z=\sqrt{2y}; dz=\frac{dz}{\sqrt{2y}}$.

Substituting these in I eventually get that $E(Z^2)=\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi}}e^{-y} \sqrt{y} \,dy = 1$, but I don't know how to bridge the gap here to $\gamma(r)=\int_{0}^{\infty}y^{r-1}e^{-y}\,dy$.

Given that it's $\Gamma\left(\frac{1}{2}\right)$, shouldn't I'm not sure how to use this in the $E(Z^2)$ equation as it's $y^{r-1}$.

Hope this was clear. Thank you!

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    $\begingroup$ I provide a statistically-motivated demonstration of this result at math.stackexchange.com/a/3558/1489 (as a byproduct of another calculation). However, the traditional approach (attributed to Liouville) is to square the integral and compute it in polar coordinates. See math.stackexchange.com/a/931944/1489 for a sketch. $\endgroup$ – whuber Mar 12 at 17:57
  • $\begingroup$ Thank you! I'll read through those proofs though I'm looking to prove it using the approach here $\endgroup$ – Sarina Mar 12 at 18:58
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    $\begingroup$ Liouville's approach is legendary as the most elementary possible solution. It's difficult to see where you can go with just elementary manipulations of the integral. But if the focus of your effort is on this particular approach, perhaps you should flag your post for migration to the math site, where there will be many more people interested in that aspect of it. $\endgroup$ – whuber Mar 12 at 19:20
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    $\begingroup$ Thank you and Liouville's approach looks like a much easier way to do it! I'll have it posted on the math site as well. Thank you $\endgroup$ – Sarina Mar 13 at 0:21
  • $\begingroup$ Please do not cross-post. It is against site rules. $\endgroup$ – StubbornAtom Mar 14 at 17:48
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Notice first

\begin{eqnarray} \Gamma(1/2) &=& \int_{0}^{\infty}y^{-1/2}e^{-y}dy \\ &=& \int_{0}^{\infty} \sqrt{2}z^{-1} e^{-z^2/2} z \ d z \qquad \text{ (substitute $y=\frac{z^2}{2}$ )} \\ &=& \frac{1}{2} \int_{-\infty}^{\infty} \sqrt{2} e^{-z^2/2} \ d z \qquad \text{ (since even function)} \\ &=& \int_{-\infty}^{\infty} \frac{1}{\sqrt{2}} z^2 e^{-z^2/2} \ d z \qquad \text{ (*using integration by parts)} \\ \end{eqnarray}

The last line (integration by parts) is valid as

\begin{eqnarray} \int_{-\infty}^{+\infty}z^{2}\left(e^{-z^2/2}\right)dz &=&\int_{-\infty}^{+\infty}z\left(ze^{-z^2/2}\right)dz \\ &=& \int_{-\infty}^{+\infty}z\left(-e^{-z^2/2}\right)'dz \\ &=& \underbrace{-ze^{-z^2/2}\Bigg|^{+\infty}_{-\infty}}_{=0}+\int_{-\infty}^{+\infty}\left(e^{-z^2/2}\right)dz\\ \end{eqnarray}

Hence we can easily see

$$\Gamma(1/2) = \frac{1}{\sqrt{\pi}} \mathbb{E}[Z^2] $$

hence

$$\Gamma(1/2) = \sqrt{\pi} \text{ □ }$$


Addition: To be clear, below I am answering exactly the question user @Sarina asked. My answer is based on a result he assumes (that the variance of the standard normal equals 1). The answer would be more complete by proving that this is the case (that involves a simple change from Cartesian to polar coordinates) but I feel this is not what user @Sarina wanted to know

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    $\begingroup$ Could you explain how the factor of $1/\sqrt{\pi}$ magically appears at the end? A look at Ben's answer might be helpful. $\endgroup$ – whuber Mar 13 at 14:34
  • $\begingroup$ Not magical. We know that $E(Z^2)=\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} z^2 e^{-z^2/2} \, dz$ and have proven that $\Gamma(1/2) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2}} z^2 e^{-z^2/2} \ d z$. Isn't is clear that $ \frac{1}{\sqrt{2\pi} = 1/\sqrt{\pi} * \frac{1}{\sqrt{\pi} $ ? $\endgroup$ – Stats Mar 13 at 15:20
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    $\begingroup$ Your "we know that" is equivalent to what the OP wishes to show. It is, as Ben wrote, "begging the question." $\endgroup$ – whuber Mar 13 at 15:26
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    $\begingroup$ Suppose you didn't know the value of $E[Z^2]$ and wished explicitly to derive it using elementary means: how would you do it? That's what the OP is asking. $\endgroup$ – whuber Mar 13 at 15:30
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    $\begingroup$ Thank you so much for this and showing how to bridge the gap between the equations!! $\endgroup$ – Sarina Mar 13 at 18:50
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This is a perfect example of a question-begging mathematical "proof": The kind of "proof" you are trying to construct here is really not terribly useful. If you don't know how to prove the integral:

$$\int \limits_0^\infty y^{-1/2} e^{-y} dy = \sqrt{\pi},$$

then you need to ask yourself why are you willing to assume that you already know the integral:

$$\int \limits_{-\infty}^\infty y^2 e^{-\tfrac{1}{2} y^2} dy = \sqrt{2 \pi},$$

which is what you are assuming when you appeal to the fact that a standard normal random variable has a variance of one. In this particular context, the properties of the normal density function first had to be derived by elementary methods, that are essentially equivalent to the result you are trying to prove. Indeed, even the fact that the normal density is a density function (i.e., that it integrates to one) is essentially just a restatement of an integral that is a simple variation of the integral you are trying to prove.

Appealing to the properties of the normal density to prove the gamma integral is an example of a "proof" that tries to demonstrate result $A$ by appealing to result $B$, where the latter theorem is a result that is itself proved by the same technique as an elementary proof of $A$. In these cases you need to be very careful because the latter result may even have a "proof" that hinges on the former result, in which case you get a completely circular argument.


Proof by elementary methods: Fortunately, in this particular case, the gamma integral and the resulting properties of the normal distribution are all provable by elementary methods. Rather than trying to "prove" the gamma integral by appealing to what are essentially variations of the same result, you can use Liouville's method (hat tip to whuber for mentioning it in comments), which uses polar coordinates to establish that:

$$\begin{equation} \begin{aligned} \Bigg( \int \limits_{-\infty}^\infty e^{- \frac{1}{2} y^2} dy \Bigg)^2 &= \Bigg( \int \limits_{-\infty}^\infty e^{- \frac{1}{2} x^2} dx \Bigg) \Bigg( \int \limits_{-\infty}^\infty e^{- \frac{1}{2} y^2} dy \Bigg) \\[6pt] &= \int \limits_{-\infty}^\infty \int \limits_{-\infty}^\infty e^{- \frac{1}{2} (x^2+y^2)} dx dy \\[6pt] &= \int \limits_0^{2 \pi} \int \limits_{0}^\infty r e^{- \frac{1}{2} r^2} dr d\theta \\[6pt] &= \int \limits_0^{2 \pi} \Bigg[ - e^{- \frac{1}{2} r^2} \Bigg]_{r=0}^{ r \rightarrow\infty} d\theta \\[6pt] &= \int \limits_0^{2 \pi} \Bigg[ 0 - (-1) \Bigg] d\theta \\[6pt] &= \int \limits_0^{2 \pi} d\theta \\[6pt] &= 2 \pi. \\[6pt] \end{aligned} \end{equation}$$

Using the substitution $r = \tfrac{1}{2} y^2$ gives $dr = y dy$, so you then have:

$$\begin{equation} \begin{aligned} \Gamma(\tfrac{1}{2}) &= \int \limits_0^\infty r^{-1/2} e^{-r} dr \\[6pt] &= \int \limits_0^\infty \sqrt{\frac{2}{y^2}} e^{- \tfrac{1}{2} y^2} y \ dy \\[6pt] &= \sqrt{2} \int \limits_0^\infty e^{- \tfrac{1}{2} y^2} \ dy \\[6pt] &= \frac{1}{\sqrt{2}} \int \limits_{-\infty}^\infty e^{- \tfrac{1}{2} y^2} \ dy \\[6pt] &= \frac{1}{\sqrt{2}} \cdot \sqrt{2 \pi} = \sqrt{\pi}. \\[6pt] \end{aligned} \end{equation}$$

This method uses only substituions and familiarity with polar coordinates, and most importantly, it does not require you to assume any premise that is effectively just a variation of the result.

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  • $\begingroup$ Thank you for the prove using Louisville's method! I prefer this way more $\endgroup$ – Sarina Mar 13 at 18:44

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