Suppose someone drops a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen. Question: What is the average of all these distances? That is to say, what is the time average of the distance traveled?
Question 1.2: Suppose someone drops a rock off a cliff of height h. As it ...
The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance will surely be less than {h}/{2} . Ignoring air resistance, the distance x at time t is x\left(t\right) = \frac{1}{2} g t^{2}.
The velocity is {dx }/{dt} = gt , and the total flight time is T = \sqrt{{2h}/{g}} .The probability that a particular photograph was taken between t and t + dt is { dt }/{T} , so the probability that it shows a distance in the corresponding range x to x + dx is
\frac{dt}{T}=\frac{dx }{gt} \sqrt{\frac{g}{2h} } = \frac{1}{2\sqrt{hx} } dx .
Thus the probability density (Equation 1.14):
{probability that an individual (chosen at random)lies between x and (x + dx)} = \rho(x)dx
is
\rho \left(x\right) = \frac{1}{2\sqrt{hx} } , ( 0\leq x \leq h )
(outside this range, of course, the probability density is zero).
We can check this result, using Equation 1.16:
\int_{-\infty}^{+\infty}{\rho(x)dx}=1\int_{0}^{h}{\frac{1}{2\sqrt{hx} }} dx = \frac{1}{2\sqrt{h} }\left(2x^{{1}/{2}} \right)\mid ^{h}_{0} = 1
The average distance (Equation 1.17):
\left\langle x\right\rangle=\int_{-\infty}^{+\infty}{\rho(x)dx}is
\left\langle x\right\rangle = \int_{0}^{h}{x\frac{1}{2\sqrt{hx} }} dx = \frac{1}{2\sqrt{h} }\left(\frac{2}{3} x^{{3}/{2}} \right)\mid ^{h}_{0} = \frac{h}{3}which is somewhat less than {h}/{2} , as anticipated.
Figure 1.7 shows the graph of \rho \left(x\right) .Notice that a probability density can be infinite though probability itself (the integral of ρ) must of course be finite (indeed, less than or equal to1).
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