I need help with this problem. Thank You The way we initially approached the Poi

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I need help with this problem.

Thank You

The way we initially approached the Poisson distribution was as an approximation: Under certain conditions where n is large enough that working directly with the Binomial distribution is unwieldy, we can work with the Poisson distribution instead and get nearly the same answer. The goal in this problem is to start getting a handle on what exactly nearly means. Consider the following three distributions: X_1 is a Binomial variable with n = 10 and p = 3/10 X_2 is a Binomial variable with n = 100 and p = 3/100 X_3 is a Binomial variable with n = 1000 and p = 3/1000 If we were to approximate these distributions by a Poisson distribution, what would the parameter m of that Poisson distribution be? Let X_0 be the Poisson distribution with your parameter m from part a. Compute P(X_1 = 0), P(X_2 = 0), P)X_3 = 0), and P(X_0 = 0) (to at least 5 digits each). If we approximate X_1 by the Poisson distribution X_0, what is the (absolute) error in our estimation of the probability that X = 0 (In other words, what is |P(X_1 = 0) - P(X_0 = 0)|)? Repeat for X_2 and X_3. What effect does multiplying n by 10 seem to have on the error in our Poisson approximation? Repeat parts b and c, but now for trying to estimate the probability that X = 6 instead of the probability that X = 0.

Explanation / Answer

Solution

Back-up Theory

1. If X is binomial with parameter n and p, then P(X = 0) = nC0p0(1 - p)n.= (1 - p)n

2. If Y is distributed as Poisson with parameter m, P(Y = 0) = (e- m)m0/0! = e- m

3. Binomial (n, p) when approximated by Poisson, the parameter of Poisson is m = np.

Now, to work out the problems,

Part (a)

In all the three cases, m = 3 [by 3 above]

Part (b)

P(X1 = 0) = (7/10)10 [by 1 above]

= 0.02825

P(X2 = 0) = (97/100)100 [by 1 above]

= 0.04755

P(X3 = 0) = (997/100)1000 [by 1 above]

= 0.04956

P(X0 = 0) = e- 3 [by 2 above]

= 0.04957

Part (c)

| P(X1 = 0) - P(X0 = 0) | = 0.02132

| P(X2 = 0) - P(X0 = 0) | = 0.00202

| P(X3 = 0) - P(X0 = 0) | = 0.00001

[The above probabilities can be easily read off from Excel]

Effect of multiplying n by 10 and dividing p by 10

Two phenomena happen simultaneously n gets 10 fold but p shrinks 10 fold making the Poisson approximation closer and closer. In mathematical language, condition for Poisson approximation to Binomial is n tends to infinity and p tends to zero which essentially means n increases while p decreases but np remains a constant.

Part (d) is identical to Part (b) and can be calculated as below:

P(X1 = 6) = 10C6 (0.3)6(0.7)4 (7/10)10 [by 1 above]

= 0.03676

P(X2 = 6) = 100C6 (0.03)6(0.97)94 [by 1 above]

= 0.04961

P(X3 = 6) = 1000C6 (0.003)6(0.997)994 [by 1 above]

= 0.05033

P(X0 = 6) = (e- 3.36)6! [by 2 above]

= 0.05041

[The above probabilities can be easily read off from Excel]

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