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SOLUTION FOR HOMEWORK 9, STAT 4351
Welcome to your 9th homework. We finish our work on classical discrete distributions.
As usual, try to find mistakes and get extra points!
Now let us look at your problems. In what follows S stands for Success.
1. Problem 5.31. Here x is a positive integer.
(a) Write using the total probability theorem:
f (x, t + δ) = P (X = x|t + δ)
= P (X = x|t)P (no S during time interval(t, t + δ))
+P (X = x − 1|t)P (1 S occurred during time interval(t, t + δ))
+
X
P (X = x − k|t)P (k Ss occurred during (t, t + δ))
k>1
= f (x, t)(1 − αδ) + f (x − 1, t)αδ.
In the last equality (which is actually an approximate equality) given information about
likelihood of occurring Ss during a small interval of time δ was used.
Now consider δ → 0 and get
df (x, t)
= α[f (x − 1, t) − f (x, t)].
dt
(b) Note that here f ∗ (x, t) = e−αt (αt)x /x!, and we can take its derivative and get:
df ∗ (x, t)
= [−αe−αt (αt)x + αe−αt x(αt)x−1 ]/x!
dt
= α[f ∗ (x − 1, t) − f ∗ (x, t)].
Please note that we deduced a nice and new approach for understanding the nature of
the Poisson RV; it can be used as a definition of Poisson RV.
2. Problem 5.38. Here I will use the following trick to avoid calculations (but you for
sure can utilize direct calculations). I may consider getting ith kind of outcome as a S and
then all other kinds of outcome as F. This yields the binomial distribution for the number
of Ss with Xi ∼ Binom(θi , n). Then E(Xi ) = nθi .
3. Problem 5.60. This is the problem on geometric distribution. Then
P (X = 6) = (.7)5 (.3)1 .
4. Problem 5.64. Here we are dealing with the hypergeometric distribution with parameters (I try to use notation of the text) N = 16, M = 10, n = 3. Let X be the number of
applicants with college degree, then:
1
(a) We write
P (X = 0) =
10! 6!
0!10! 3!3!
16!
3!13!
P (X = 1) =
10! 6!
1!9! 2!4!
16! .
3!13!
P (X = 2) =
10! 6!
2!8! 1!5!
16! .
3!13!
P (X = 3) =
10! 6!
3!7! 0!6!
16! .
3!13!
.
(b)
(c)
(d)
5. Problem 5.68. Here we are dealing with the hypergeometric distribution having
parameters N = 80, M = 4 and n = 3. Let X be the number of defective alarms in the
shipment. Then:
(a)
P (X = 1) =
4! 76!
1!3! 2!74!
80!
3!77!
.
(b) For this approximation n = 3, θ = 4/80 = .05, and then
P (Xb = 1) =
3!
(.05)1 (.95)2 .
1!2!
6. Problem 5.73. The underlying problem is clearly on Binomial distribution with Xb ∼
Binom(p = .014, n = 150). Because the probability p of a single S in the Bernoulli trial
is very small and E(Xb ) = np = 2.1, we can approximate the Binomial distribution by the
Poisson with the intensity (mean) λ = np = 2.1. [Note that this is the case when you can
say that the Poisson distribution describes a rare event. At the same time, if np ≥ 5 we
can use normal approximation as well. In practice, you should look at how well this or that
approximation fits your setting; there exist special methods to do this —- we may discuss
them in the statistical part of the course.]
Then
P (Xpois = 2) = e−λ λ2 /2! = e−2.1 (2.1)2 /2.
7. Problem 5.74. Here again the underlying random variable of interest is Xb ∼
Binom(p = .0012, n = 1000). Because p is small and n is moderately large, a Poisson
approximation looks reasonable (at least you may try to use it and compare with the binomial — now-a-day it is easy with the help of computers). Consider Xpois ∼ P oisson(λ = np).
Here λ = np = (1000)(.0012) = 1.2. Note that for the Normal approximation we need np ≥ 5
[as a rule of thumb], and we are not there yet with the considered sample size. Then
P (Xbin ≤ 2) ≈ e−λ [λ0 /0! + λ1 /1! + λ2 /2!].
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8. Problem 5.81. This is not a simple problem (as one can think) because we need to
understand the random variable in question. It is given that the number of imperfections
per yard is X1 ∼ P ois(λ = .25). Then it is reasonable to assume that the number of
imperfections in another yard is an independent Poisson random variable X2 with the same
intensity λ = .25. Then the number of imperfections per two yards is Y = X1 + X2 . Now we
need to understand what is the distribution of Y . Recall that the mgf (moment generating
function) of X1 is
t
MX1 (t) = E(etX1 ) = eλ(e −1) .
Here the mgf is unique description of the RV, so by looking at the mgf of Y ,
MY (t) = E(etY ) = E(et(X1 +X2 ) ) [now I use independence]
t
t
t
E(etX1 )E(etX2 ) = eλ(e −1) eλ(e −1) = e2λ(e −1) ,
we conclude that Y ∼ P ois(2λ). In other words, we get that the sum of two independent
and identical Poisson RVs is again Poisson with the doubled intensity [please note that a
”natural” feeling Y = 2X1 does not imply the desired result].
We may conclude that
P (X1 + X2 ≤ 1) = P (Y ≤ 1) = e−2λ [(2λ)0 /0! + (2λ)1 /1!].
9. Problem 5.83. Given: n = 10, θ1 = .4, θ2 = .5, θ3 = .1 and Y = (X1 , X2 , X3 ) ∼
Multinomial(n, θ1 , θ2 , θ3 ). Then
P (X1 = 3, X2 = 6, X3 = 1) =
10! 3 6 1
θ θ θ .
3!6!1! 1 2 3
10. Problem 5.87. Given: N = 25, M1 = 15, M2 = 7, M3 = 3, n = 5. Here we are
dealing with the multivariate hypergeometric distribution.
(a)
P (X1 = 4, X2 = 1, X3 = 0) =
15! 7! 3!
4!11! 1!6! 0!3!
25!
5!20!
.
P (X1 = 3, X2 = 1, X3 = 1) =
15! 7! 3!
3!12! 1!6! 1!2!
25!
5!20!
.
(b)
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