Mean and Variance of DistributionsDate: 12/07/2000 at 09:37:18 From: Vaughn Duff Subject: Mean and Variance of Distributions I'm stuck on the following problems. Any insight into deriving a solution would be greatly appreciated. Thank you! I am trying to derive the expected value (mean) and variance for these distributions: 1) Exponential with parameter 1/k: f(w) = (k)*e^(-w*k) or maybe it should be f(w)=(1/k)*e^(-w/k) The expected value is equal to: SUM[f(w)*w] (discrete case) or inf INT[f(w)*w dw] (continuous case) 0 The variance is equal to: SUM[f(w)*w^2 dw - (mean)^2] The derivation should be equal to: E(w) = 1/k var(w) = 1/k^2 2) Hypergeometic distribution: define y = the number of S's in a random sample of size n (taken without replacement) from a population of N_s S's and N_f F's: p(y) = ((N_s C y)*(N_f C n-y))/(N C n) The definition of the mean and variance are the same as in the first case. The derivation should be equal to: E(y) = n*(N_s/N) var(y) = n*(N_s/N)*(1-N_s/N)*((N-n)/(N-1)) Date: 12/07/2000 at 11:03:04 From: Doctor Anthony Subject: Re: Mean and Variance of Distributions For the first case, show that for a Poisson process the time interval between events has an exponential distribution. If the average is k per unit time, then in t units of time the average would be kt. The probability of no event P(0) in time t is therefore e^(-kt). The probability P(T>t) = prob. no events in time 0 -> t = e^(-kt), so P(T<t) = 1 - e^(-kt) This is the c.d.f. of the time to first event. To get the p.d.f. we differentiate this and have: f(t) = ke^(-kt) This is pdf of time interval between events, and is the exponential distribution. inf E(t) = INT[t.ke^(-kt).dt] 0 and integrating by parts = t(-1/k)ke^(-kt) + INT[e^(-kt).dt] = 0 - (1/k)[e^(-kt)] from 0 to infinity = -(1/k)[0 - 1] E(t) = 1/k To get variance we first find: E(t^2) = INT[t^2.(ke^(-kt).dt] and integrating by parts (twice) gives 2/k^2 and so var(t) = 2/k^2 - 1/k^2 = 1/k^2 For the second case, suppose we have a population of b black and g green elements. A sample of size r is taken (without replacement) and we wish to find the expected number of black elements in the sample. We could calculate this by direct methods but a better way is to introduce a new variable x(k) which takes the value 1 or 0 depending on whether the kth element in the sample is or is not black. By symmetry this probability for each value of k is b/(b+g) The expected number of black elements for each k is: E[x(k)] = 1.b/(b+g) + 0.g/(b+g) = b/(b+g) E[b] = E[x(1)] + E[x(2)] + ... + E[x(r)] and so E[b] = b/(b+g) + b/(b+g) + ... + b/(b+g) to r terms giving: E[b] = rb/(b+g) Variance -------- To calculate the variance we continue as follows: E(x(k)^2) = 1^2.b/(b+g) + 0.g/(b+g) = b/(b+g) Var(x(k)) = b/(b+g) - b^2/(b+g)^2 b(b+g) - b^2 = ------------ (b+g)^2 bg = ------- (b+g)^2 Now we must add together the means and variances of the individual results of the x(i)'s. To do this we first require the covariance of x(j).x(k) Covariance ----------- x(j).x(k) = 1 if BOTH x(j) and x(k) = 1, otherwise x(j).x(k) = 0. Probability x(j).x(k) = 1 is given by: b b-1 --- . ----- b+g b+g-1 and so E[x(j).x(k)] = b(b-1)/[(b+g)(b+g-1)] COV[x(j).x(k)] = E[x(j).x(k)] - E(x(j)).E(x(k)) = b(b-1)/[(b+g)(b+g-1)] - [b/(b+g)].[b/(b+g)] b(b-1)(b+g) - b^2(b+g-1) = ------------------------ (b+g)^2.(b+g-1) b^2(b+g) - b(b+g) - b^2(b+g) + b^2 = ---------------------------------- (b+g)^2.(b+g-1) -b^2 - bg + b^2 = --------------- (b+g)^2.(b+g-1) -bg = --------------- (b+g)^2.(b+g-1) And so finally for a sample size r taken from the population of b+g we get: rb E[S(r)] = --- b+g The covariance of x(j) with all other x(i) gives r-1 of these terms, so: rbg r-1 Var[S(r)] = ------- [1 - -----] (b+g)^2 b+g-1 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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