Sums of Random Variables n Consider n RVs x i and let s x i . n i - PowerPoint PPT Presentation

Sums of Random Variables n Consider n RVs x i and let s ≡ x i . n i =1 If the RVs are statistically independent, then < s > = n < x i > i Var( s ) = n Var( x i ) i 8.044 L4B1

• The individual p ( x i ) could be quite different • Both continuous and discrete RVs could be present • True for any n • Even if one RV dominates the sum 8.044 L4B2

Results have a special meaning when 1) The means are finite ( = 0) 2) The variances are finite ( = ∞ ) 3) No subset dominates the sum 4) n is large p(s) width 1 ∝ mean n ∝ n s ∝ n 8.044 L4B3

Given p ( x, y ), find p ( s ≡ x + y ) y A x+y = α α y = α− x α x dx � α − ζ � ∞ B P s ( α ) = dη p x,y ( ζ, η ) dζ −∞ −∞ 8.044 L4B4

� ∞ C p s ( α ) = dζ p x,y ( ζ, α − ζ ) −∞ This is a general result; x and y need not be S.I. Application to the Jointly Gaussian RVs in Section 2 shows that p ( s ) is a Gaussian with zero mean and a Variance = 2 σ 2 (1 + ρ ). 8.044 L4B5

In the special case that x and y are S.I. � ∞ dζ p x ( ζ ) p y ( α − ζ ) = � ∞ dζ I p x ( α − ζ I ) p y ( ζ I ) p s ( α ) = −∞ −∞ The mathematical operation is called “convolu- tion”. � ∞ p ( z ) q ( x − z ) dz = f ( x ) . p ⊗ q ≡ −∞ 8.044 L4B6

Example p(z) Given: ∝ (z/a)n e -z/a 1 ( z/a ) n exp( − z/a ) p ( z ) = n ! a ∝ z n 1 ( z/a ) m exp( − z/a ) q ( z ) = z m ! a 0 < z and n, m = 0 , 1 , 2 , · · · Find: p ⊗ q 8.044 L4B7

q(z) q(-z) z z 0 0 p(z) q(x-z)=q(-(z-x)) q(x-z) x z x z 0 0 finite product � 8.044 L4B8

n m � � � � 1 1 z x − z x − z/a e − ( x − z ) /a dz p ⊗ q = e 2 0 n ! m ! a a a n + m +1 � � 1 1 1 x − x/a z n ( x − z ) m dz = e n ! m ! a 0 a n + m +1 � � 1 1 x 1 − x/a ζ n (1 − ζ ) m dζ = e n ! m ! a 0 a � n ! m ! ( n + m +1)! 8.044 L4B9

� � n + m +1 1 1 x − x/a p ⊗ q = e ( n + m + 1)! a a a function of the same class 8.044 L4B10

Example Atomic Hydrogen Maser RF out flask ν 0 ν = 1.4....... GHz H* beam ν 1 −ν 0 about 10 KHz ν 1 p(t wall | n stays) = ? cavity 8.044 L4B11 �

n t wall (given n stays) = n t i i =1 t i ≡ duration of i th stay on wall. Each stay is S.I. − t/τ p ( t | 1) = (1 /τ ) e − t/τ p ( t | 2) = p ( t | 1) ⊗ p ( t | 1) = (1 /τ )( t/τ ) e p ( t | 3) = p ( t | 2) ⊗ p ( t | 1) = (1 / 2)(1 /τ )( t/τ ) 2 e − t/τ 8.044 L4B12

� � n − 1 1 1 t − t/τ p ( t | n ) = e ( n − 1)! τ τ τ p(t | 12 ) 0.15 0.12 0.10 0.1 0.08 0.06 0.05 0.04 0.02 t / τ 5 10 15 20 25 30 0 10 20 30 8.044 L4B13

Facts about sums of RVs • Exact expressions for < s > and Var( s ) if S.I. • p ( s ) = p ( x ) ⊗ p ( y ) if S.I. • p ( s ) slightly more complicated if not S.I. 8.044 L4B14

• ⊗ usually changes functional form • But not always • Fourier techniques are very useful 8.044 L4B15

Very important special case: Central Limit Theorem • RVs are S.I. • All have identical densities p ( x i ) • Var( x ) is finite but < x > could be zero • n is large p(s) Central Limit Theorem: p(s) is Gaussian ∝ n s ∝ n 8.044 L4B16

If x is continuous √ 1 − ( s − <s> ) 2 / 2 σ 2 p ( s ) = e 2 πσ 2 < s > = n < x > σ 2 = n σ 2 x 8.044 L4B17

If x is discrete in equal steps of ∆ x ∆ x − ( s − <s> ) 2 / 2 σ 2 p ( s ) = e δ ( s − i ∆ x ) √ 2 πσ 2 v ) i comb v ) envelope p(s) s ∆ x 8.044 L4B18

Non­rigorous extensions of the Central Limit Theorem • The Gaussian can be a good practical approxima- tion for modest values of n . • The Central Limit Theorem may work even if the individual members of the sum are not identically distributed. • The requirement that the variables be statistically independent may even be waived in some cases, particularly when n is very large 8.044 L4B19

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