Suppose X_1.....X_n are IID^6 rv\'s with common CDF Pois- son Po(X). Please resp

Question

Suppose X_1.....X_n are IID^6 rv's with common CDF Pois- son Po(X). Please respond to: Calculate the mean and variance of X_1 +...X_n. Which is the exact CDF of X_1 +...X_n? Which property can you use to solve this question'? How could you approximate this CDF for large n, using an appropriate normal CDF? How could you approximate the CDF of X = 1/n(X_1 + +X_n), for large, n, using an appropriate, normal CDF? Simulate 100 IID sample of (X_1, X_n), for n = 2. Obtain and graph the histogram of X = 1/n(X_1+ +X_n) Repeat the previous question for n = 5, 10 and 20. Compare the histograms. What can you conclude?

Explanation / Answer

Back-up Theory

If a continuous random variable, X, is uniformly distributed over the interval (a, b), then the pdf (probability density function) of X is given by

f(x) = 1/(b – a) …………………….…………………………………………………………(1)

CDF (cumulative distribution function) = P(X t) = Integral (a to t) of f(x) = (t – a)/(b – a).(2)

Mean or Expected Value, E(X) = (a + b)/2 ………………………………………………….(3)

Variance, V(X) = (b - a)2/12…………….. ………………………………………………….(4)

Standard Deviation, SD(X) = (b - a)/(23)…………………………………………………..(5)

If X1, X2, X3, ……. , Xn are iid with mean µ and variance 2, then

E(X1 + X2 + X3 + …….+ Xn) = nµ ……………………………………………(6)

V(X1 + X2 + X3 + …….+ Xn) = n2 ……………………………………………(7)

E(Xbar) = µ ……………………………………………………………………….(8)

V(Xbar) = 2/n ……………………………………………………………………….(9)

Now, to work out solution,

In the given question, X ~ U(0, 1), i.e., a = 0 and b = 1. So, [vide (3), (4) and (5) under Back-up Theory], E(X) = ½, V(X) = 1/12 and SD(X) = 1/(23)

Part (1)

[vide (6) and (7) under Back-up Theory],

Mean of (X1 + X2 + X3 + …….+ Xn) = n/2 and

Variance of (X1 + X2 + X3 + …….+ Xn) = n/12 ANSWER

Part (2)

Let S = (X1 + X2 + X3 + …….+ Xn). Then, by Central Limit Theorem, for large n,

CDF of [{S – E(S)}/SE(S)] can be approximated by CDF of N(0,1).

So, CDF of S = P(S < s) = P[Z < {s – (n/2)}/(n/12)]

i.e., CDF of S = P(S < x) = [{x – (n/2)}/(n/12)] ANSWER [ represents CDF of N(0, 1)]

Part (3)

Identical to the above, CDF of Xbar = P(Xbar < x) = [{x – (1/2)}/(1/12n)] ANSWER [ represents CDF of N(0, 1)]

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