Question

Q4. (20pts, Binomial and Poisson approximation Suppose a gambler bets (1) ten times on events of probability 1/20, (2) then twenty times on events of probability 1/20, (3) then thirty times on events of probability 1/30, (4) then forty times on events of probability 1/40. Assuming the events are independent. (i) What is the exact distribution of the number of times the gambler wins in (4)? (It suffices to say the name of the distribution with appropriate parameter(s).) (ii) What is the approximate distributions of the number of times the gambler wins in (0)?

Answer 1

(i)
The exact distribution of the number of times the gambler wins in (4) is Binomial distribution with parameters n = 40, p = 1/40

(ii)
The Binomial distribution can be approximated as Poisson distribution with parameter $\lambda$ = np when

np < 10 for large n and small values of p.

np = 40 * (1/40) = 1 which is less than 10.

Thus, the approximate distribution of the number of times the gambler wins in (4) is Poisson distribution with parameter $\lambda$ = 1

(iii)

MGF of X1 and X2 is,

$M_{X_1}(t) = e^{\mu_1(e^t-1)}$

$M_{X_2}(t) = e^{\mu_2(e^t-1)}$

MGF of X1 + X2 is,

$M_{X_1 + X_2}(t) = M_{X_1}(t) .M_{X_2}(t) = e^{\mu_1(e^t-1)} . e^{\mu_2(e^t-1)} = e^{(\mu_1 + \mu_2)(e^t-1)}$

which is MGF of Poisson distribution with parameter $\lambda$ = $\mu_1 + \mu_2$

Thus, X1 + X2 ~ Poisson($\lambda$ = $\mu_1 + \mu_2$ )

(iv)

Let X1, X2, X3 and X4 be the distribution of (1), (2), (3) and (4) respectively.

Then, the exact distributions of X1, X2, X3 and X4 are,

X1 ~ Binomial(n = 10, p = 1/10)

X2 ~ Binomial(n = 20, p = 1/20)

X3 ~ Binomial(n = 30, p = 1/30)

X4 ~ Binomial(n = 40, p = 1/40)

The approximate distributions of X1, X2, X3 and X4 are,

X1 ~ Poisson($\lambda$ = 1)

X2 ~ Poisson($\lambda$ = 1)

X3 ~ Poisson($\lambda$​​​​​​​ = 1)

X4 ~ Poisson($\lambda$​​​​​​​ = 1)

where $\lambda$ = np

By (iii), X1 + X2 + X3 + X4 ~ Poisson($\lambda$ = 1+ 1+ 1+ 1 = 4)

X1 + X2 + X3 + X4 ~ Poisson($\lambda$ = 4)

The approximate distribution of the total number of wins of all four bets is Poisson($\lambda$ = 4)