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The density of a continuous random variable $X$ is

$$f_X(x) = \begin{cases}\frac{p+1}{2}|x|^p&\mathrm{\ if\ } |x|\le1\\ 0&\mathrm{\ otherwise }\end{cases}$$

Here $p$ is a parameter taking values between $-0.5$ and $0.5$. How do I find the values of $p$ that maximize and minimize $Var(X)$?

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2 Answers 2

up vote 3 down vote accepted

Doing it from first principles, you have

$$\mathrm{Var}(X)=\mathrm{E}\left[\left(X-\mathrm{E}(X)\right)^2\right]=\int_{-\infty}^\infty(x-\mu)^2f_X(x)~dx\;,$$ where $$\mu=\int_{-\infty}^\infty xf_X(x)~dx\;.$$

The function $f_X(x)$ is even (i.e., symmetric about the $y$-axis), so

$$\begin{align*} \int_{-\infty}^0 xf_X(x)~dx&=\int_\infty^0(-x)f_X(-x)~d(-x)\\ &=\int_0^\infty(-x)f_X(x)~dx\\ &=-\int_0^\infty xf_X(x)~dx\;, \end{align*}$$

and therefore $$\mu=\int_{-\infty}^0 xf_X(x)~dx+\int_0^\infty xf_X(x)~dx=0\;.$$ Thus,

$$\begin{align*} \mathrm{Var}(X)&=\int_{-1}^1x^2\left(\frac{p+1}2\right)|x|^p~dx\\ &=\frac{p+1}2\int_{-1}^1 x^2|x|^p~dx\\ &=(p+1)\int_0^1 x^{p+2}~dx\\ &=\frac{p+1}{p+3}\;. \end{align*}$$

Now it’s just a first-semester calculus problem in maximizing and minimizing the function $$v(p)=\frac{p+1}{p+3}=1-\frac2{p+3}$$ over the interval $\left[-\frac12,\frac12\right]$. This is easy: $$v'(p)=\frac2{(p+3)^2}>0$$ over the entire interval, so the variance is increasing over the entire interval. Thus, it must have its minimum value at $p=-\frac12$ and its maximum at $p=\frac12$. (Actually, you shouldn’t even need any calculus to see that $v(p)$ is increasing.)

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Shouldn't it be $\int_{-\infty}^{\infty}(x−μ)^2fX(x) dx$ and not $\int_{-\infty}^{\infty}(x−μ)fX(x) dx$? –  idealistikz Oct 1 '12 at 3:42
@idealistikz: It should indeed; thanks. –  Brian M. Scott Oct 1 '12 at 5:19

Given that $|x|^p$ has even symmetry, we conclude that $\mathbb{E} (X) = 0$. Therefore, the variance is

$$\text{Var} (X) = \mathbb{E} (X^2) = \displaystyle\int_{-1}^1 x^2 f_X (x) dx = \displaystyle\int_{-1}^1 x^2 \left(\frac{p+1}{2}\right) |x|^p dx = \left(\frac{p+1}{2}\right) \displaystyle\int_{-1}^1 x^2 |x|^p dx$$

and, since the integrand has even symmetry, we obtain

$$\text{Var} (X) = (p+1) \displaystyle\int_{0}^1 x^{p+2} dx = \left(\frac{p+1}{p+3}\right)$$

Assuming that this is correct, then plot the graph of $f (p) := \frac{p+1}{p+3}$ to optimize the variance.

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Plotting the graph is both unnecessary and an uncertain method. –  Brian M. Scott Sep 30 '12 at 8:59
@BrianM.Scott: True. But I am lazy and felt like shortening the post. –  Rod Carvalho Sep 30 '12 at 9:03
As Brian says: $\dfrac{p+1}{p+3} = 1-\frac{2}{p+3}$ which makes it easy and short –  Henry Sep 30 '12 at 15:48
@Henry: Leaving the "easy and short" work to the OP is even easier and shorter. Unless I am paid by the word, which I am not, I will take the liberty of leaving some work to the OP. –  Rod Carvalho Sep 30 '12 at 17:21

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