Computational Physics - Department of Physics

(Axel Boer) #1

354 11 Outline of the Monte Carlo Strategy


p(x) =
λx
x!
e−λ x= 0 , 1 ,...,;λ> 0.

In this case both the mean value and the variance are easier tocalculate,


μ=



x= 0

x
λx
x!
e−λ=λe−λ



x= 1

λx−^1
(x− 1 )!
=λ,

and the variance isσ^2 =λ. An example of applications of the Poisson distribution could be the
counting of the number ofα-particles emitted from a radioactive source in a given timeinter-
val. In the limit ofn→∞and for small probabilitiesy, the binomial distribution approaches
the Poisson distribution. Settingλ=ny, withythe probability for an event in the binomial
distribution we can show that


nlim→∞

(

n
x

)

yx( 1 −y)n−xe−λ=



x= 1

λx
x!e

−λ,

see for example Refs. [63,64] for a proof.


11.2.1Multivariable Expectation Values


An important quantity is the so called covariance, a variantof the variance. Consider the
set{Xi}ofnstochastic variables (not necessarily uncorrelated) withthe multivariate PDF
P(x 1 ,...,xn). Thecovarianceof two of the stochastic variables,XiandXj, is defined as follows


Cov(Xi,Xj)≡


(xi−〈xi〉)(xj−〈xj〉)


=


···


(xi−〈xi〉)(xj−〈xj〉)P(x 1 ,...,xn)dx 1 ...dxn (11.8)

with
〈xi〉=



···


xiP(x 1 ,...,xn)dx 1 ...dxn

If we consider the above covariance as a matrixCi j=Cov(Xi,Xj), then the diagonal elements
are just the familiar variances,Cii=Cov(Xi,Xi) =Var(Xi). It turns out that all the off-diagonal
elements are zero if the stochastic variables are uncorrelated. This is easy to show, keeping in
mind the linearity of the expectation value. Consider the stochastic variablesXiandXj, (i 6 =j)


Cov(Xi,Xj) =


(xi−〈xi〉)(xj−〈xj〉)


=〈xixj−xi〈xj〉−〈xi〉xj+〈xi〉〈xj〉〉
=〈xixj〉−〈xi〈xj〉〉−〈〈xi〉xj〉+〈〈xi〉〈xj〉〉
=〈xixj〉−〈xi〉〈xj〉−〈xi〉〈xj〉+〈xi〉〈xj〉
=〈xixj〉−〈xi〉〈xj〉

IfXiandXjare independent, we get〈xixj〉=〈xi〉〈xj〉, resulting inCov(Xi,Xj) = 0 (i 6 =j).
Also useful for us is the covariance of linear combinations of stochastic variables. Let{Xi}
and{Yi}be two sets of stochastic variables. Let also{ai}and{bi}be two sets of scalars.
Consider the linear combination


U=∑
i

aiXi V=∑
j

bjYj
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