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There is a youtube video dealing with the proof that the sums of the squares of normally distributed $n$ random errors, each one distributed as $\sim \chi^2(1\text{ df})$ follows a chi square distribution with $n$ degrees of freedom:

So if $X_1, \ldots, X_n$ are $\sim \chi^2 \,\small(1\text{ df})$, then $X_1 +\cdots+X_n \sim \chi^2\,\small(n\text{ df})$.

I got stuck in one step, which is highlighted below, and wonder if someone could explain. I thought the derivation was great, and I'm transcribing it (sorry about the errors) with the idea that someone may be looking for something like this on-line (like I was for quite some time) without much luck (again the step I am inquiring about is in bold):


Proof through the $cdf$:

$p\,(X_1 +\cdots+X_n \leq x)= p\,(Y_1^2 +\cdots+Y_n^2 \leq x)$, where $X_i =Y _i^2$, and $Y \sim N(0,1)$.

$Y_1^2 +\cdots+Y_n^2 \leq x$ is an $n$-dimensional sphere with radius $\,\sqrt{x}$: $Y_1^2 +\cdots+Y_n^2 \leq \sqrt[]{x^2}$.

So the probability is:

$p\,(X_1 +\cdots+X_n \leq x) = \displaystyle \int\cdots\int_{(Y_1^2 +\cdots+Y_n^2 \leq x)}\large f_{\small(Y_1,\ldots,Y_n)}\large(y_1,\ldots,y_n)\, dy_1,\ldots,dy_n$ with $f$ being the joint density. Remembering that they are iid rv's and that $Y$ is normal:

$$p(X_1 +\cdots+X_n \leq x) = \displaystyle \int\cdots\int_{(Y_1^2 +\cdots+Y_n^2 \leq x)} \bigg(\frac{1}{2\pi}\bigg)^n e^{-\bigg(\frac{y_1^2}{2}+\cdots + \frac{y_n^2}{2}\bigg) } dy_1\cdots dy_n$$

Changing to spherical coordinates it will be something like:

$dy_1\cdots dy_n=dr\,r\,d\theta_1\,\ldots\,r\,d\theta_n \cdot g(\theta_1\,\ldots\,\theta_n)$ with $g$ being some function of the angles.

And, $$\small p(X_1 +\cdots+X_n \leq x) = \displaystyle \int_{r=0}^{\sqrt{x}} \int_{\theta_1=\alpha_1}^{\beta_1} \ldots\int_{\theta_{n-1}=\alpha_{n-1}}^{\beta_{n-1}} \bigg(\frac{1}{2\pi}\bigg)^{n/2}\, e^{-(\frac{r^2}{2}) } \,r^{n-1} \,g(\theta_1\,\ldots\,\theta_{n-1})\, dr\,d\theta_1\cdots \,d\theta_n.$$

There is no $r$ dependency in the $\theta$ integration. The integration of the $\theta$'s gets absorbed into a constant that depends on $n$, $C_n$:

$$p(X_1 +\cdots+X_n \leq x) = C_n \cdot \displaystyle \int_{r=0}^{\sqrt{x}} e^{-\bigg(\frac{r^2}{2}\bigg) } r^{n-1} \, dr.$$

This is the distribution ($cdf$), and to get the density ($pdf$) we have to take derivatives:

Let's $f(t)=e^{-({t^2}/2)} \,t^{n-1}$, such that $F'(t)=f(t)$.

$$F_{X_1,\ldots,X_n}(x)=C_n\cdot F(\sqrt{x})$$

$$f_{X_1,\ldots,X_n}(x)=C_n\cdot f(\sqrt{x})\frac{1}{2\sqrt{x}}$$

$$f_{X_1,\ldots,X_n}(x)=C_n\cdot e^{-x/2}\,x^{\frac{n-1}{2}}\,\frac{1}{2}\,x^{-\frac{1}{2}}= \frac{C_n}{2}\cdot e^{-x/2}\,x^{\frac{n}{2}-1},$$

which is a chi square with n degrees of freedom if we equate $\frac{C_n}{2}=\frac{1}{2^{n/2}\Gamma(\frac{n}{2})}$ given that the pdf of chi-square is:

$$\displaystyle f_X(x)= \frac{1}{2^{\nu/2}\,\Gamma(\frac{\nu}{2})}\,e^{-x/2}\,x^{\frac{\nu}{2}-1}$$ for $x\leq 0$. Otherwise, $0$.

Geometrically, this is equivalent to the surface area of the $n$-dimensional sphere.


How can you seemingly whimsically decide that the derivative of your $cdf$ will conveniently be $f(t)=e^{-(t^2/2)} \,t^{n-1}$? It's like we didn't need a proof if we were going to pull this one out of thin air... (I'm sure there's an explanation).

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    $\begingroup$ One of the forms of the Fundamental Theorem of Calculus deals with the derivative of a function whose value is the value of an integral, and the function argument appears in the limits, If $$F(\alpha)=\int_{a(\alpha)}^{b(\alpha)}f(x)\mathrm dx$$, then $$\frac{\mathrm dF(\alpha)}{\mathrm d\alpha}=f(b(\alpha))\frac{\mathrm db(\alpha)}{\mathrm d\alpha}-f(a(\alpha))\frac{\mathrm db(\alpha)}{\mathrm d\alpha}.$$ So, apply this result. $\endgroup$ – Dilip Sarwate Aug 4 '15 at 3:17
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    $\begingroup$ Using Leibniz rule $$\frac{d}{dx}\left[C_n \displaystyle \int_{r=0}^{\sqrt{x}} e^{-\bigg(\frac{r^2}{2}\bigg) } r^{n-1} dr\right]=C_n \frac{1}{2\sqrt{x}}e^{-x/2}x^{(n-1)/2}$$ $\endgroup$ – d.k.o. Aug 4 '15 at 3:25
  • $\begingroup$ I think a simpler approach is to use moment generating functions. $\endgroup$ – BruceET Aug 4 '15 at 6:43

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