Let N be Poisson-distributed with mean lambda. So. P(N=n) = lambda n/n!e-lambda, n = 0, 1, 2, 3, Compute Var(x)
Solution
var(X)=E(X^2)-(E(X))^2 E(X2) = ?k=0 k^2*(1/k!)*?^k*e-? ------>Definition of Poisson distribution Note change of limit: term is zero when k=0 = ?*e-?[?k=1(k/(k-1)!)*?^k-1 = ?e-?(?k=1 ((k-1)/(k-1)!)*?^(k-1) +?k=1 (1/(k-1)!)*?^(k-1)) straightforward algebra Again, note change of limit: term is zero when k-1=0 =?e-???*?k=2 (1/(k-2)!)*?^(k-2) +?k=1(1(k-1)!)*?^(k-1)? = ?e-???*?i=0 (1/i!)*?^i+?j=0 (1/j!)*?^j? --> putting i=k-2,j=k-1 = ?e-?(?*e^?+e*?) Taylor Series Expansion for Exponential Function = ?(?+1) = ?^2+? Then: var(X) = E(X^2)-(E(X))^2 = ?^2+?-?^2 , Expectation of Poisson Distribution: E(X)=? = ?
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Let N be Poisson-distributed with mean lambda- So- P(N-n) - lambda n-.docx
1. Let N be Poisson-distributed with mean lambda. So. P(N=n) = lambda n/n!e-lambda, n = 0, 1, 2,
3, Compute Var(x)
Solution
var(X)=E(X^2)-(E(X))^2 E(X2) = ?k=0 k^2*(1/k!)*?^k*e-? ------>Definition of Poisson
distribution Note change of limit: term is zero when k=0 = ?*e-?[?k=1(k/(k-1)!)*?^k-1 = ?e-
?(?k=1 ((k-1)/(k-1)!)*?^(k-1) +?k=1 (1/(k-1)!)*?^(k-1)) straightforward algebra Again, note
change of limit: term is zero when k-1=0 =?e-???*?k=2 (1/(k-2)!)*?^(k-2) +?k=1(1(k-1)!)*?^(k-
1)? = ?e-???*?i=0 (1/i!)*?^i+?j=0 (1/j!)*?^j? --> putting i=k-2,j=k-1 = ?e-?(?*e^?+e*?) Taylor
Series Expansion for Exponential Function = ?(?+1) = ?^2+? Then: var(X) = E(X^2)-(E(X))^2 =
?^2+?-?^2 , Expectation of Poisson Distribution: E(X)=? = ?
