70 Chapter 3. The molecular dance[[Student version, December 8, 2002]]
Let us work out the variance of the Gaussian distribution, Equation 3.8. Changing variables as
in the Example on page 68, we need to compute
variance(x)=
2 σ^2
√
π∫∞
−∞dyy^2 e−y
2. (3.12)
Todo this we need a trick, which we’ll use again later: Define a functionI(b)by
I(b)=∫∞
−∞dye−by
2Again changing variables givesI(b)=
√
π/b.Nowconsider the derivative dI/db.Onone hand it’sdI/db=−1
2
√
π
b^3. (3.13)On the other hand,
dI/db=∫∞
−∞dyd
db
e−by
2
=−∫∞
−∞dyy^2 e−by
2. (3.14)
Settingb=1,that last integral is the one we needed (see Equation 3.12)! Combining Equations 3.13,
3.14, and 3.12 gives variance(x)=^2 σ
√^2
π
(
−ddIb∣∣
b=1)
=^2 σ
√^2
π×√π
2 .Thusthe RMS deviation of the
Gaussian distribution just equals the parameterσappearing in Equation 3.8.
3.1.4 Addition and multiplication rules
Addition rule Section 3.1.1 noted that for a discrete distribution, the probability that the next
measured value ofxis eitherxiorxjequalsP(xi)+P(xj), unlessi=j.The key thing is that
xcan’t equalbothxiandxj;wesaythe alternative values areexclusive. More generally, the
probability that a person is either taller than 2mor shorter than 1. 9 mis obtained by addition,
whereas the probability to be either taller than 2mor nearsighted cannot be obtained in this way.
Foracontinuous distribution, the probability that the next measured value ofxis either between
aandbor betweencanddequals the sum,
∫b
adxP(x)+∫d
cdxP(x), provided the two intervals
don’t overlap. That’s because the two probabilities (to be betweenaandbor betweencandd)are
exclusive in this case.
Multiplication rule Now suppose we measure two independent quantities, for example, tossing
acoin and rolling a die. What is the probability that we get headsandroll a 6? To find out, just
list all 2×6=12 possibilities. Each is equally probable, so the chance of getting the specified one
is 1/12. This shows that the joint probability distribution for two independent events is theproduct
of the two simpler distributions. LetPjoint(xi,yK)bethe joint distribution wherei=1or 2 and
x 1 =“heads,”x 2 =“tails”; similarlyyK=K,whereK=1,...,6isthe number on the die. Then
the multiplication rule says
Pjoint(xi,yK)=Pcoin(xi)×Pdie(yK). (3.15)Actually Equation 3.15 is correct even for loaded dice (thePdie(yK)aren’t all equal to^16 )ora
two-headed coin (Pcoin(x 1 )=1,Pcoin(x 2 )=0). On the other hand, for two connected events (for
example, the chance of rain versus the chance of hail) we don’t get such a simple relation.
Your Turn 3c
Show that ifPcoinandPdieare correctly normalized, then so will bePjoint.