9. Line Integrals

c. Applications of Scalar Line Integrals

3. Probability & Expected Value

There is also a probability interpretation of weighted averages:

Let \(p(x,y,z)\) be a function defined along the curve \(\vec r(t)=(x(t),y(t),z(t))\) between \(A=\vec r(a)\) and \(B=\vec r(b)\). If \(\displaystyle \int_A^B p\,ds=1\), then \(p(x,y,z)\) is called a probability distribution along the curve. If \(C\) and \(D\) are points on the curve between \(A\) and \(B\), then \(\displaystyle P(C,D)=\int_C^D p\,ds\) is interpreted as the probability we have selected a point on the curve between \(C\) and \(D\). Further, the expected value of a function \(f(x,y,z)\) is \[ \langle f\rangle =\int_A^B f\,p\,ds =\int_a^b f(\vec r(t))\,p(\vec r(t))|\vec v|\,dt \]

An expected value is actually a weighted average with weight \(p\) because \(\displaystyle \int_A^B p\,ds=1\). Specifically, \[ f_{p\text{-ave}} =\dfrac{\displaystyle \int_A^B f\,p\,ds} {\displaystyle \int_A^B p\,ds} =\displaystyle \int_A^B f\,p\,ds =\langle f\rangle \]

1. Find \(a\) so that \(p(x,y)\) is a probability distribution.
2. What is the probability that the oxygen molecule is between \(x=\dfrac{1}{2}\) and \(x=1\)?
3. What is the expected value of \(y\) for the oxygen molecule?

The curve may be parametrized by \(\vec r(t)=(t,t^2)\). So its velocity is \(\vec v=(1,2t)\) whose length is \(|\vec v|=\sqrt{1+4t^2}\).

1. On the curve \(p=ax=at\). To be a probability distribution, we need \(\displaystyle \int_{(0,0)}^{(\sqrt{2},2)} p\,ds=1\). However, (with \(u=1+4t^2\)) \[\begin{aligned} \int_{(0,0)}^{(\sqrt{2},2)} &p\,ds =\int_0^{\sqrt{2}} at\,\sqrt{1+4t^2}\,dt =\dfrac{a}{8}\int_1^9 \sqrt{u}\,du \\ &=\left[\dfrac{au^{3/2}}{12}\right]_1^9 =\left(\dfrac{a27}{12}\right)-\left(\dfrac{a}{12}\right) =\dfrac{13a}{6}=1 \end{aligned}\] So we need \(a=\dfrac{6}{13}\). Then \(p=\dfrac{6}{13}x=\dfrac{6}{13}t\).
2. The probability that the oxygen molecule is between \(x=\dfrac{1}{2}\) and \(x=1\) is: \[\begin{aligned} P\left(\dfrac{1}{2},1\right)&=\int_{(1/2,1/4)}^{(1,1)} p\,ds =\int_{1/2}^1 \dfrac{6}{13}x|\vec v|\,dt \\ &=\int_{1/2}^1 \dfrac{6}{13}t\,\sqrt{1+4t^2}\,dt =\dfrac{6}{13}\dfrac{1}{8}\int_2^5 \sqrt{u}\,du \\ &=\dfrac{3}{52}\left[\dfrac{2}{3}u^{3/2}\right]_2^5 =\dfrac{1}{26}\left(5^{3/2}-2^{3/2}\right) \approx.32 \end{aligned}\]
3. The expected value of \(y\) is \[\begin{aligned} \langle y\rangle&=\int_{(0,0)}^{(\sqrt{2},2)} y\,p\,ds =\int_0^{\sqrt{2}} t^2\dfrac{6}{13}t\,\sqrt{1+4t^2}\,dt \\ &=\dfrac{6}{13}\dfrac{1}{8}\int_1^9 \left(\dfrac{u-1}{4}\right)\sqrt{u}\,du \\ &=\dfrac{3}{208}\int_1^9 \left(u^{3/2}-u^{1/2}\right)\,du =\dfrac{3}{208}\left[\dfrac{2u^{5/2}}{5}-\,\dfrac{2u^{3/2}}{3}\right]_1^9 \\ &=\dfrac{3}{208}\left(\dfrac{2\cdot243}{5}-\,\dfrac{2\cdot27}{3}\right) -\,\dfrac{3}{208}\left(\dfrac{2}{5}-\,\dfrac{2}{3}\right) \\ &=\dfrac{149}{130} \approx1.15 \end{aligned}\]

Consider the function \(p(x,y,z)=axy\) along the twisted cubic \(\vec r(t)=\left\langle t,t^2,\dfrac{2}{3}t^3\right\rangle\) between \(t=0\) and \(t=1\).

1. Find \(a\) so that \(p\) is a probability distribution.

   \(\displaystyle a=\dfrac{12}{7}\) and \(p=\dfrac{12}{7}xy=\dfrac{12}{7}t^3\).

   On the curve, \(p=axy=at^3\) and \(|\vec v|=1+2t^2\). To be a probability distribution, we need \(\displaystyle \int_{\vec r(0)}^{\vec r(1)} p\,ds=1\). Specifically: \[\begin{aligned} \int_{\vec r(0)}^{\vec r(1)} p\,ds &=\int_0^1 at^3(1+2t^2)\,dt =a\left[\dfrac{t^4}{4}+\dfrac{t^6}{3}\right]_0^1 \\ &=a\left(\dfrac{1}{4}+\dfrac{1}{3}\right) =\dfrac{7}{12}a=1 \end{aligned}\] So \(a=\dfrac{12}{7}\) and \(p=\dfrac{12}{7}xy=\dfrac{12}{7}t^3\).

2. What is the probability that we select a point on the curve with \(0 \le t \le \dfrac{1}{2}\)?

   \(\displaystyle P\left(0,\dfrac{1}{2}\right)=\dfrac{1}{28} \approx 0.0357\)

   The probability that we select a point on the curve with \(0 \le t \le \dfrac{1}{2}\) is: \[\begin{aligned} P\left(0,\dfrac{1}{2}\right)&= \int_{\vec r(0)}^{\vec r(1/2)} p\,ds =\int_0^{1/2} \dfrac{12}{7}t^3(1+2t^2)\,dt \\ &=\dfrac{12}{7}\left[\dfrac{t^4}{4}+\dfrac{2t^6}{6}\right]_0^{1/2} =\dfrac{12}{7}\left(\dfrac{1}{2^4 4}+\dfrac{2}{2^6 6}\right) \\ &=\dfrac{12}{2^6 7}\dfrac{4}{3} =\dfrac{1}{28} \approx 0.0357 \end{aligned}\]

3. What is the expected value of \(z\) for the curve?

   \(\displaystyle \langle z\rangle =\dfrac{184}{441} \approx 0.417\)

   The expected value of \(z\) is \[\begin{aligned} \langle z\rangle&= \int_{\vec r(0)}^{\vec r(1)} z\,p\,ds =\int_0^1 \dfrac{2}{3}t^3\dfrac{12}{7}t^3(1+2t^2)\,dt \\ &=\dfrac{8}{7}\left[\dfrac{t^7}{7}+\dfrac{2t^9}{9}\right]_0^1 =\dfrac{8}{7}\left(\dfrac{1}{7}+\dfrac{2}{9}\right) \\ &=\dfrac{184}{441} \approx 0.417 \end{aligned}\]