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26 Probability would be if you’d just taken a single measurement. Of course, you may still have systematic errors in your experiment. If you are consistently overestimating your quantity by an error in your experimental setup, that error won’t reduce by repeated measurement! A second application is in the theory of random walks. Imagine a drunken person staggering out of a pub and attempting to walk along a narrow street (which confines him or her to motion in one dimen- sion). Let’s pretend that with each inebriated step, the drunken person is equally likely to travel one step forward or one step backward. The effects of intoxication are such that each step is uncorrelated with the previous one. Thus the average distance travelled in a single step is X = 0. After n such steps, we would have an expected total distance travelled of Y = X i = 0. However, in this case the root mean squared distance is more revealing. In this case Y 2 = nX 2 , so that the rms length of a random walk of n steps is n times the length of a single step. This result will be useful in considering Brownian motion in Chapter 33. Chapter summary In this chapter, several introductory concepts in probability theory have been introduced. The mean of a discrete probability distribution is given by x = i x i P i , and the mean of a continuous probability distribution is given by x = xP (x)dx. The variance is given by σ 2 x = (x x) 2 , where σ x is the standard deviation. If y = ax + b, then y = ax + b and σ y = x . If u and v are independent random variables, then uv = uv. In particular, if Y = X 1 + X 2 + ··· + X n , where the X’s are all from the same distribution, Y = nx and σ Y = X .

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26 Probability

would be if you’d just taken a single measurement. Of course, you maystill have systematic errors in your experiment. If you are consistentlyoverestimating your quantity by an error in your experimental setup,that error won’t reduce by repeated measurement!

A second application is in the theory of random walks. Imagine adrunken person staggering out of a pub and attempting to walk alonga narrow street (which confines him or her to motion in one dimen-sion). Let’s pretend that with each inebriated step, the drunken personis equally likely to travel one step forward or one step backward. Theeffects of intoxication are such that each step is uncorrelated with theprevious one. Thus the average distance travelled in a single step is〈X〉 = 0. After n such steps, we would have an expected total distancetravelled of 〈Y 〉 =

∑〈Xi〉 = 0. However, in this case the root meansquared distance is more revealing. In this case 〈Y 2〉 = n〈X2〉, so thatthe rms length of a random walk of n steps is

√n times the length of

a single step. This result will be useful in considering Brownian motionin Chapter 33.

Chapter summary

• In this chapter, several introductory concepts in probability theoryhave been introduced.

• The mean of a discrete probability distribution is given by

〈x〉 =∑

i

xiPi,

and the mean of a continuous probability distribution is given by

〈x〉 =

∫xP (x) dx.

• The variance is given by

σ2x = 〈(x− 〈x〉)2〉,

where σx is the standard deviation.

• If y = ax+ b, then 〈y〉 = a〈x〉 + b and σy = aσx.

• If u and v are independent random variables, then 〈uv〉 = 〈u〉〈v〉.In particular, if Y = X1 + X2 + · · · + Xn, where the X’s are allfrom the same distribution, 〈Y 〉 = n〈x〉 and σY =

√nσX .

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Further reading 27

Further reading

There are many good books on probability theory and statistics. Recommended ones include Papoulis (1984), Walland Jenkins (2003) and Sivia and Skilling (2006).

Exercises

(3.1) A throw of a regular die yields the numbers 1, 2,. . . , 6, each with probability 1/6. Find the mean,variance and standard deviation of the numbers ob-tained.

(3.2) The mean birth weight of babies in the UK is about3.2 kg with a standard deviation of 0.5 kg. Convertthese figures into pounds (lb), given that 1 kg =2.2 lb.

(3.3) This question is about a discrete probability distri-bution known as the Poisson distribution. Letx be a discrete random variable which can take thevalues 0, 1, 2, . . . A quantity is said to be Poissondistributed if one obtains the value x with proba-bility

P (x) =e−mmx

x!,

where m is a particular number (which we will showin part (b) of this exercise is the mean value of x).

(a) Show that P (x) is a well-behaved probabilitydistribution in the sense that

∞Xx=0

P (x) = 1.

(Why is this condition important?)

(b) Show that the mean value of the probability

distribution is 〈x〉 =∞X

x=0

xP (x) = m.

(c) The Poisson distribution is useful for describingvery rare events which occur independentlyand whose average rate does not change overthe period of interest. Examples include birthdefects measured per year, traffic accidents ata particular junction per year, numbers of ty-pographical errors on a page, and number ofactivations of a Geiger counter per minute.The first recorded example of a Poisson dis-tribution, the one which in fact motivated

Poisson, was connected with the rare eventof someone being kicked to death by a horsein the Prussian army. The number of horse-kick deaths of Prussian military personnel wasrecorded for each of 10 corps in each of 20years from 1875–1894 and the following datarecorded:

Number of deaths Observedper year frequencyper corps

0 1091 652 223 34 1

≥ 5 0

Total 200

Calculate the mean number of deaths peryear per corps. Compare the observed fre-quency with a calculated frequency assumingthe number of deaths per year per corps arePoisson distributed with this mean.

(3.4) This question is about a continuous probability dis-tribution known as the exponential distribution.Let x be a continuous random variable which cantake any value x ≥ 0. A quantity is said to be ex-ponentially distributed if it takes values between xand x + dx with probability

P (x) dx = Ae−x/λ dx

where λ and A are constants.

(a) Find the value of A that makes P (x) a well-defined continuous probability distribution so