Answer:
Variance Σ i=1 (1 - Pi) /Pi^2
Expected value= π(1/1 + 1/2 .......... 1/n) = π.Hn
Step-by-step explanation:
From the information given, we can get
E(T) = E(t1) + E(t2) ......... +E(tn)
1/P1 + 1/P2.......... 1/Pn
-π/r + π/ r-1 +........... π/1
π(1/1 + 1/2 .......... 1/n)
To find the variance,
Variance= var (x1 + x2......... +xr)
E i=1 var (Xi)
Where,
x1, x2, x3 ........... are all independent of each other.
In case of xi : var (xi) = E(xi^2) - E(xi^2)
Probability of the ith term of coupon that would be observed
Pi = (n - i - 1)/n
Therefore,
i-1 coupons out of a total of n coupons.
To calculate Exi, probability Pi
The expected number of coupons required should be = 1
In probability, 1 - Pi expected number of coupons required E (Xi +1)
E(Xi) = Pi + (1- Pi) E(xi + 1)
E(Xi) = 1/Pi
Due to the above,
E(Xi)^2= Pi + (1- Pi) E(xi + 1)^2
E(Xi)^2= Pi + (1- Pi) E(xi^2 + 1 + 2xi)
E(Xi)^2= 2/Pi^2 - 1/Pi
Var (Xi) = 2/Pi^2 - 1/Pi - 1/Pi^2
= (1 - Pi) /Pi^2
Variance Σ i=1 (1 - Pi) /Pi^2
