Week 8 Activity

Section 8.2 Week 8 Activity

Subsection 8.2.1 Probability Distributions

Activity 8.2.1.

Consider this probability distribution defined on the interval \([0,6]\text{.}\)

\begin{equation*} f(x) = a(6x - x^2) \end{equation*}

Determine the value of the constant \(a\) that makes this a probability distribution.
Find the probability of a measurement in \([1,3]\text{.}\)
Find the probability of a measurement in \([2,4]\text{.}\)
Calculate the mean.
Calculate the standard deviation.

Solution.

Figure 8.2.1.

I integrate the function over the domain \([0,6]\text{.}\)

\begin{equation*} \int_0^6 f(x) dx = \int_0^6 a(6x-x^2) = a \left( 3x^2 - \frac{x^3}{3} \Bigg|_0^6 \right) = a(108 - 72) = 36a \end{equation*}

This integral needs to be \(1\text{.}\) That happens when \(a = \frac{1}{36}\text{.}\) I’ll use this value of \(a\) for the rest of the question. The resulting function is shown in Figure 8.2.1.
I integrate the function over \([1,3]\) to find the probability of a measurement in that range.

\begin{align*} P([1,3]) \amp = \int_1^3 f(x) dx = \frac{1}{36} \int_1^3 6x-x^2 dx = \frac{1}{36} 3x^2 - \frac{x^3}{3} \Bigg|_1^3 \\ \amp = \frac{1}{36} \left[ 27-9-3+\frac{1}{3} \right] = \frac{46}{108} = \frac{23}{54} \end{align*}
I integrate the function over \([2,4]\) to find the probability of a measurement in that range.

\begin{align*} P([2,4]) \amp = \int_2^4 f(x) dx = \frac{1}{36} \int_2^4 6x-x^2 dx = \frac{1}{36} 3x^2 - \frac{x^3}{3} \Bigg|_2^4 \\ \amp = \frac{1}{36} \left[ 48 - \frac{64}{3} - 12 + \frac{18}{3} \right] = \frac{52}{108} = \frac{13}{27} \end{align*}
I integrate \(xf(x)\) to calculate the mean.

\begin{align*} \mu \amp = \int_0^6 x f(x) dx = \frac{1}{36} \int_0^6 6x^2 - x^3 = \frac{1}{36} \left[ 2x^3 - \frac{x^4}{4} \Bigg|_0^6 \right]\\ \amp = \frac{1}{36} \left[ 432 - \frac{1296}{4} \right] = \frac{1}{36} [ 432 - 324] = \frac{108}{36} = 3 \end{align*}
I integrate \((x-\mu)^2 f(x)\) to calculate \(\sigma^2\text{.}\) Then I take the square root to calculate \(\sigma\text{.}\)

\begin{align*} \sigma^2 \amp = \int_0^6 (x-3)^2 f(x) dx = \frac{1}{36} \int_0^6 (x^2-6x+9)(6x-x^2) dx \\ \amp = \frac{1}{36} \int_0^6 -x^4 + 12x^3 - 45 x^2+ 54x dx \\ \amp = \frac{1}{36} \left[ \frac{-x^5}{5} + 3x^4 - 15x^3 + 27x^2 \Bigg|_0^6 \right] \\ \amp = \frac{1}{36} \left[ \frac{-7776}{5} + 3888 - 3240 + 972 \right] \\ \amp = \frac{-7776 + 19440 - 16200 + 4860}{180} = \frac{324}{180} = \frac{9}{5}\\ \sigma \amp = \sqrt{\frac{9}{5}} \doteq 1.34 \end{align*}

Activity 8.2.2.

Consider this probability distribution defined on the interval \([0,6]\text{.}\)

\begin{equation*} g(x) = \left\{ \begin{matrix} ax \amp x \in [0,3] \\ a(6-x) \amp x \in [3,6] \end{matrix} \right. \end{equation*}

Determine the value of the constant \(a\) that makes this a probability distribution.
Find the probability of a measurement in \([1,3]\text{.}\)
Find the probability of a measurement in \([2,4]\text{.}\)
Calculate the mean.
Calculate the standard deviation.

Solution.

Figure 8.2.2.

I integrate the function over the domain \([0,6]\text{.}\) Since this is a piecewise function, the integral splits into two integrals for the two pieces of the domain.

\begin{align*} \int_0^6 g(x) dx \amp = \int_0^3 x dx + \int_3^6 (6-x) dx = \frac{x^2}{2} \Bigg|_0^3 + 6x - \frac{x^2}{2} \Bigg|_3^6\\ \amp = \frac{9}{2} + 36 - 18 - 18 + \frac{9}{2} = 9 \end{align*}

This integral needs to be \(1\text{.}\) That happens when \(a = \frac{1}{9}\text{.}\) I’ll use this value of \(a\) for the remainder of the question. This function is show in Figure 8.2.2.
I integrate the function over \([1,3]\) to find the probability of a measurement in that range.

\begin{equation*} P([1,3]) = \int_1^3 g(x) dx = \int_1^3 \frac{x}{9} dx = \frac{x^2}{18} \Bigg|_1^3 = \frac{9}{18} - \frac{1}{9} = \frac{8}{18} = \frac{4}{9} \end{equation*}
I integrate the function over \([2,4]\) to find the probability of a measurement in that range.

\begin{align*} P([2,4]) \amp = \int_2^4 g(x) dx = \int_2^3 \frac{x}{9} dx + \int_3^4 \frac{6-x}{9} dx \\ \amp = \frac{x^2}{18} \Bigg|_2^3 + \frac{6x}{9} - \frac{x^2}{18} \Bigg|_3^4 \\ \amp = \frac{9}{18} - \frac{4}{18} + \frac{24}{9} - \frac{16}{18} - \frac{18}{9} + \frac{9}{18} = \frac{10}{18} = \frac{5}{9} \end{align*}
I integrate \(xg(x)\) to calculate the mean.

\begin{align*} \mu \amp = \int_0^6 x g(x) dx = \int_0^3 \frac{x^2}{9} dx + \int_3^6 \frac{6x-x^2}{9} dx = \frac{x^3}{27} \Bigg|_0^3 + \frac{6x^2}{18} - \frac{x^3}{27} \Bigg|_3^6 \\ \amp = \frac{27}{27} - 0 + \frac{216}{18} - \frac{216}{27} - \frac{54}{18} + \frac{27}{27} = 1 + 12 - 8 - 3 + 1 = 3 \end{align*}
I integrate \((x-\mu)^2 g(x)\) to calculate \(\sigma^2\text{.}\)

\begin{align*} \sigma^2 \amp = \int_0^6 (x-3)^2 g(x) dx \\ \amp = \int_0^3 (x^2-6x+9)\frac{x}{9} dx + \int_3^6 (x^2-6x+9)\frac{(6-x)}{9} dx \\ \amp = \frac{1}{9} \int_0^3 x^3 - 6x^2 + 9x dx + \frac{1}{9} \int_3^6 -x^3 +12x^2 - 45 x + 54 dx \\ \amp = \frac{1}{9} \left[ \frac{x^4}{4} - 2x^3 + \frac{9x^2}{2} \Bigg|_0^3 + \frac{-x^4}{4} + 4x^3 - \frac{45x^2}{2} + 54 x \Bigg|_3^6 \right] \\ \amp = \frac{1}{9} \left[ \frac{81}{4} - 54 + \frac{81}{2} - \frac{1296}{4} + 864 - \frac{1620}{2} \right. \\ \amp \left. + 324 + \frac{81}{4} - 108 + \frac{405}{2} - 162 \right]\\ \amp = \frac{1}{36} \left[ 81 - 216 + 162 - 1296 + 3456 - 3240 + 1296 \right. \\ \amp \left. + 81 - 432 + 810 - 648 \right]\\ \amp = \frac{54}{36} = \frac{3}{2} \\ \sigma \amp = \sqrt{\frac{3}{2}} \doteq 1.22 \end{align*}

Activity 8.2.3.

Give a reason why it is reasonable that the standard deviation of Activity 8.2.1 is larger than the standard deviation of Activity 8.2.2, even though they have the same domain and the same mean.

Solution.

The first distribution is much broader than the second. It extends wider and only decays at the very edge of its domain. Therefore, we expect that typical measurements are farther from the mean and the standard deviation is higher.

Activity 8.2.4.

Consider this probability distribution defined on the domain \([1, \infty)\text{,}\) where \(r\) is a parameter with \(r \gt 1\text{.}\)

\begin{equation*} f(x) = \frac{r-1}{x^r} \end{equation*}

Find the probability of a measurement in \([1,2]\) (this will depend on the parameter \(r\) of course).
Try to calculate the mean. Argue that this calculation is only possible when \(r \gt 2\text{.}\)
For \(r \in (1, 2]\text{,}\) this is a probability distribution which has no calculatable mean. How can this be so?
Try to calculate the standard deviation. Conclude that the standard deviation is only calculatable when \(r \gt 3\text{.}\)
For \(r \in (2,3]\text{,}\) this is a probability distribution with a mean but not calculatable standard deviation. How can this be so?

Solution.

To get the probability, I integrate the probability distribution over the range \([1,2]\text{.}\)

\begin{equation*} P([1,2]) = (r-1)\int_1^2 \frac{1}{x^r} dx = (r-1) \frac{1}{x^{r-1}}\frac{1}{1-r} \Bigg|_1^2 = 1- \frac{1}{2^{r-1}} \end{equation*}
I try to calculate the integral of \(xf(x)\) to calculate the mean.

\begin{equation*} \mu = (r-1)\int_1^\infty x \frac{1}{x^r} dx = (r-1)\int_1^\infty \frac{1}{x^{r-1}} dx \end{equation*}

If \(r \in (1,2]\text{,}\) then this integrand is either \(\frac{1}{x}\) (when \(r=2\) or something larger. The integral of \(\frac{1}{x}\) is \(\ln x\text{.}\) Evaluated on these limits, the improper integral will diverge. When \(r \lt 2\text{,}\) since the integral is even large than \(\frac{1}{X}\text{,}\) the improper integral will likewise diverge.

However, when \(r>2\text{,}\) the improper integral will converge. I’ll do that calculation now.

\begin{align*} \mu \amp = (r-1)\int_1^\infty x \frac{1}{x^r} dx = (r-1)\int_1^\infty \frac{1}{x^{r-1}} dx \\ \amp = (r-1)\lim_{a \rightarrow \infty} \frac{1}{x^{r-2}}\frac{1}{2-r} \Bigg|_1^a\\ \amp = (r-1)\lim_{a \rightarrow \infty} \frac{1}{(2-r)(a^{r-2})} + \frac{1}{r-2} = \frac{r-1}{r-2} \end{align*}
The tail is simply to long. Even though the function decays and has finite area, the long tail skews the average higher and higher. For \(r \lt 2\text{,}\) the effect of that skew is enough to make the mean meaningless.
The argument is the same as for the mean. Since I need to square \((x-\mu)\text{,}\) we will have an \(x^2\) term in the numerator. That means the integral will have a \(\frac{1}{x^{r-2}}\) term, which won’t lead to a finite integral unless \(r \gt 3\text{.}\) If \(r \gt 3\text{,}\) I’ll do the calculation of the standard deviation.

\begin{align*} \sigma^2 \amp = (r-1)\int_1^\infty \left( x - \frac{r-1}{r-2} \right)^2 \frac{1}{x^r} dx \\ \amp = (r-1) \int_1^\infty \frac{x^2}{x^r} - 2 \left( \frac{r-1}{r-2} \right) \frac{x}{x^r} + \left( \frac{r-1}{r-2} \right)^2 \frac{1}{x^r} dx\\ \amp = (r-1) \int_1^\infty \frac{1}{x^{r-2}} - 2 \left( \frac{r-1}{r-2} \right) \frac{1}{x^{r-1}} + \left( \frac{r-1}{r-2} \right)^2 \frac{1}{x^r} dx\\ \amp = (r-1) \left[ \frac{1}{x^{r-3}}\frac{1}{3-r} - 2 \left( \frac{r-1}{r-2} \right) \frac{1}{x^{r-2}} \frac{1}{2-r} \right.\\ \amp + \left. \left( \frac{r-1}{r-2} \right)^2 \frac{1}{x^{r-1}} \frac{1}{1-r} \Bigg|_1^\infty \right]\\ \amp = (r-1) \lim_{a \rightarrow \infty} \left[ \frac{1}{x^{r-3}}\frac{1}{3-r} - 2 \left( \frac{r-1}{r-2} \right) \frac{1}{x^{r-2}} \frac{1}{2-r} \right.\\ \amp \left. + \left( \frac{r-1}{r-2} \right)^2 \frac{1}{x^{r-1}} \frac{1}{1-r} \Bigg|_1^a \right]\\ \amp = (r-1) \left[ \frac{1}{r-3} - 2\left( \frac{r-1}{r-2} \right) \frac{1}{r-2} + \left( \frac{r-1}{r-2} \right)^2 \frac{1}{r-1} \right]\\ \amp = \left( \frac{r-1}{r-3} \right) -2 \left( \frac{r-1}{r-2} \right)^2 + \left( \frac{r-1}{r-2} \right)^2 = \left( \frac{r-1}{r-3} \right) - \left( \frac{r-1}{r-2} \right) \end{align*}
This is the same situation as for the mean. The long tail also skews the standard deviation. When \(r \leq 3\text{,}\) this skew is enough to send the standard deviation to infinity.

Subsection 8.2.2 Conceptual Review Questions

What is a probability distribution?
Why is the measurement the independent variable for a probability distribution?
Why do integrals calculate probabilities, means and standard deviations?
What is a central tendency? Why are there multiple mathematical definition of a central tendency?

Prev Top Next