10 h FIG.:The probability density in the Examplep() 2点水Temh:bt and6ae Pu=de)de and the ruleswe deduced for discrete distributions translate in the obvious way: (回=xpe)da Uen=CIee达 a2=(△x〉-x)-2 it falls:i distance ut beeth Inongin he distancetime e time near the top,so the average x0=)9t2 片e共 要-√景=后“ Evidently the probability density is 网-2症0≤rs) 1 。2后-a(n0-110 FIG. 8: The probability density in the Example ρ(x). ρ(x) is the probability of getting x, or probability density. The probability that x lies between a and b (a finite interval) is given by the integral of ρ(x) Pab = ∫ b a ρ(x)dx and the rules we deduced for discrete distributions translate in the obvious way: 1 = ∫ +∞ −∞ ρ(x)dx ⟨x⟩ = ∫ +∞ −∞ xρ(x)dx ⟨f(x)⟩ = ∫ +∞ −∞ f(x)ρ(x)dx σ 2 ≡ ⟨ (∆x) 2 ⟩ = ⟨ x 2 ⟩ − ⟨x⟩ 2 Example: Suppose I drop 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 distance? That is to say, what is the time average of the distance traveled? Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance must be less than h/2. Ignoring air resistance, the distance x at time t is x(t) = 1 2 gt2 The velocity is dx/dt = gt, and the total flight time is T = √ 2h/g. The probability that the camera flashes in the interval dt is dt/T, so the probability that a given photograph shows a distance in the corresponding range dx is dt T = dx gt √ g 2h = 1 2 √ hx dx Evidently the probability density is ρ(x) = 1 2 √ hx ,(0 ≤ x ≤ h) (outside this range, of course, the probability density is zero.) We can check the normalization of this result ∫ h 0 1 2 √ hx dx = 1 2 √ h ( 2x 1/2 )h 0 = 1