Finding the Maximum of the Chi-Square density function

probabilityprobability distributionsproof-verification

Below is a problem which I did and I believe my answer is correct. I would like
somebody to confirm that (if true) and provide some additional comments about my style.
Thanks,
Bob
Problem:
Show that $\chi^2_v$ pdf has a maximum at $v – 2$ if $v > 2$.
Answer:
The $\chi^2_v$ pdf for $x >= 0$ is:
\begin{eqnarray*}
f_(x) &=& \begin{cases}
\frac{ x^{\frac{v}{2} – 1}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)}
& \text{for } x >= 0 \\
0 & \text{otherwise} \\
\end{cases} \\
\end{eqnarray*}
\newline
Now to find its maximum we compute $f'(x)$.
\begin{eqnarray*}
f'(x) &=&
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} – 1}e^{-\frac{x}{2}} + ( \frac{v}{2} – 1) x^{\frac{v}{2} – 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)} \\
\end{eqnarray*}
Now we set $f'(x) = 0$.
\begin{eqnarray*}
\frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} – 1}e^{-\frac{x}{2}} + ( \frac{v}{2} – 1) x^{\frac{v}{2} – 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)} &=& 0 \\
\big(-\frac{1}{2}\big)x^{\frac{v}{2} – 1}e^{-\frac{x}{2}} + ( \frac{v}{2} – 1) x^{\frac{v}{2} – 2}e^{-\frac{x}{2}} &=& 0 \\
\big(-\frac{1}{2}\big)x^{\frac{v}{2} – 1} + ( \frac{v}{2} – 1) x^{\frac{v}{2} – 2} &=& 0 \\
\big(-\frac{1}{2}\big)x + \frac{v}{2} – 1 &=& 0 \\
x &=& v – 2 \\
\end{eqnarray*}
Now we know that $x = v – 2$ is an extreme point. The question is it a maximum
or a minimum. Since $f(0) = 0$ and $f(v-2) > 0$ we conclude that $x = v – 2$ is a maximum.

Best Answer

What you did makes a lot of sense, but what if it $v-2$ was a saddle point? A better way to see it is using the double derivative test, i.e.

\begin{equation} f'(x) = \frac{\big(-\frac{1}{2}\big)x^{\frac{v}{2} - 1}e^{-\frac{x}{2}} + ( \frac{v}{2} - 1) x^{\frac{v}{2} - 2}e^{-\frac{x}{2}}} {2^{\frac{v}{2}} \Gamma(v/2)} \end{equation} Then \begin{equation} f''(x) = \frac{1}{2^{\frac{v}{2}} \Gamma(v/2)} \dfrac{x^{\frac{v}{2}-3}\left(x^2+\left(4-2v\right)x+v^2-6v+8\right)\mathrm{e}^{-\frac{x}{2}}}{4} \end{equation} At the extremum, we have \begin{equation} f''(v - 2) = \frac{(v-2)^{\frac{v}{2}-3}}{4(2^{\frac{v}{2}} \Gamma(v/2))} ( (v-2)^2 -2(v-2)(v-2) + (v-2)(v-4) )e^{-\frac{v-2}{2}} \end{equation} that is \begin{equation} f''(v - 2) = \frac{(v-2)^{\frac{v}{2}-3}e^{-\frac{v-2}{2}}}{4(2^{\frac{v}{2}} \Gamma(v/2))} ( -(v-2)^2 + (v-2)(v-4) ) \end{equation} which is \begin{equation} f''(v - 2) = -2 \frac{(v-2)^{\frac{v}{2}-3}e^{-\frac{v-2}{2}}}{4(2^{\frac{v}{2}} \Gamma(v/2))} (v-2) \end{equation} For $v > 2$ it is easy to see that $f''(v-2) < 0$, hence it is a maximum.

Related Solutions

[Math] Change of variable on a probability density function – $\sin(x)$

We know that: $$f_y(y) = f_y(x) \left| \frac{dx}{dy} \right|$$

This part is wrong. This holds only when the change of variables is invertible. And your function $x \mapsto \sin x$ is definitely not invertible on $(0,2\pi)$.

In this case you may want to get into the basic principles. The p.d.f. of a continuous random variable is defined as the derivative of the c.d.f. of $X$

$$ f_X(x) = \frac{d}{dx}\mathbb{P}(X \leq x) $$

and we know how to deal with c.d.f. In our case, $Y$ can take values in $[-1,1]$ so it is only meaningful to investigate the range $y \in [-1, 1]$. Then

$$\mathbb{P}(Y \leq y) = \mathbb{P}(\sin X \leq y) = \frac{2\pi - 2\arccos y}{2\pi}. $$

(If you can't see this easily, consider two cases $y \in [0, 1]$ and $y \in [-1, 0]$ separately and try to read out the above probability from the graph of the sine function.) So we have

$$ f_Y(y) = \frac{d}{dy}\mathbb{P}(Y \leq y) = \frac{1}{\pi\sqrt{1-y^2}} $$

when $y \in [-1, 1]$ and $f_Y(y) = 0$ outside $[-1, 1]$.

[Math] An Example dealing with the Chi-Square Test

With a $\chi^2$ test we reject if our test statistic is greater than the critical value.

Let $$H_0: p=0.5$$

$$H_a: p\neq0.5$$

We have $$\chi^2=\sum \frac{(O-E)^2}{E}$$

where $E=30$ for both groups so

$$\begin{align*} \chi^2 &=\sum \frac{(O-E)^2}{E}\\\\ &=\frac{(37-30)^2}{30}+\frac{(23-30)^2}{30}\\\\ &\approx3.27 \end{align*}$$

We look at a $\chi^2$ table at $\alpha=0.05$ with $df=1$ and get a critical value of $3.84$. Thus we fail to reject the null hypothesis since $3.27\lt 3.84$.

We look at a $\chi^2$ table at $\alpha=0.01$ with $df=1$ and get a critical value of $6.635$. Thus we fail to reject the null hypothesis since $3.27\lt 6.635$.

Using software you should find

1 - pchisq(3.27, 1)
[1] 0.07055762

so we fail to reject at either significance level since $0.706\gt0.05\gt0.01$

Another Test:

Using a binomial test, the probability of getting $23$ or less heads or $23$ or less tails when flipping a fair coin $60$ times is

$$2\cdot\sum_{k=0}^{23} {60 \choose k}0.5^{60}\approx0.0925$$

> sum(dbinom(0:23,60,.5))*2
[1] 0.09246098

so using this test, we fail to reject at both significance levels as well.

Best Answer

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[Math] Change of variable on a probability density function – $\sin(x)$

[Math] An Example dealing with the Chi-Square Test

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