ISyE 3232 Stochastic Manufacturing and Service Models Fall 2015

Z. She & YL. Chang

Homework 1 - SolutionsAugust 21, 2015

1. (a) Poisson Distribution

(b) Normal Distribution

(c) Binomial Distribution

(d) Bernoulli Distribution

(e) Geometric Distribution

(f) Exponential Distribution

2. We are given that E[X] = 3 and V ar(X) = 20. Thus;

(a) E[5− 2X − 6X] = E[5− 8X] = 5− 8E[X] = 5− 24 = −19V ar(5− 2X − 6X) = V ar(5− 8X) = 82V ar(X) = 64V ar(X) = 1280

(b) E[(3X − 4)/3 + 5] = E[X + 11/3] = E[X] + 11/3 = 3 + 11/3 = 20/3V ar((3X − 4)/3 + 5) = V ar(X + 11/3) = V ar(X) = 20

3. Let’s say h denotes the number of heads and t denotes the number of tails obtained when a coin istossed 5 times. Then, h+ t = 5We want to find the difference d between them: d = h− tWe know that h = 5− t. So, d = h− t = 5− 2tThe number of tailes t ∈ {0, 1, 2, 3, 4, 5}Then, 5− 2t = d ∈ {−5,−3,−1, 1, 3, 5}

4. The distribution function of X is given by

F (b) =

0 b < 0

1/3 0 ≤ b < 12/5 1 ≤ b < 47/8 4 ≤ b < 4.51 b ≥ 4.5

We need calculate the probability mass function of X, so we need to find P (X = i) for i = 0, 1, 4, 4.5F (b) = P (X ≤ b)P (X = i) = P (X ≤ i)− P (X < i) = F (i)− P (X < i) which is the jump between the values of b.

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

P (X = 0) = F (0)− P (X < 0) =1

3− 0 =

1

3

P (X = 1) = F (1)− P (X < 1) =2

5− 1

3=

1

15

P (X = 4) = F (4)− P (X < 4) =7

8− 2

5=

19

40

P (X = 4.5) = F (4.5)− P (X < 4.5) = 1− 7

8=

1

8

The probability mass function of X is given by

p(i) =

1/3 i = 01/15 i = 119/40 i = 41/8 i = 4.50 otherwise

5. Pr{X = i} = ci for positive, even integers i ≤ 11

(a) Let’s say i = 2k where k = 1, 2, 3, 4, 5We know that

1 =∑5k=0 P (X = 2k) =

∑5k=0 2ck = 2c

5× (5 + 1)

2= 30c

Then, c=1/30

(b) E[X] =∑5k=0 2kP (X = 2k) =

∑5k=0(2k)(2ck) =

∑5k=0 c(2k)2 = 4c

∑5k=0 k

2

= 4/30(12 + 22 + 32 + 42 + 52) = 22/3

(c) E[X2] =∑5k=0(2k)2P (X = 2k) =

∑5k=0(2k)2(2ck) =

∑5k=0 c(2k)3 = 8c

∑5k=0 k

3

= 8/30(13 + 23 + 33 + 43 + 53) = 60

(d) V ar(X) = E[X2]− (E[X])2 = 60− (22/3)2 = 56/9

(e) Note that (7−X)+ is defined as max(7−X, 0). Then,

(7−X)+ =

5 with probability 2c (when X = 2)3 with probability 4c (when X = 4)1 with probability 6c (when X = 6)0 with probability 18c (when X = 8, 10)

Therefore, E[(7−X)+] = 5× 2c+ 3× 4c+ 1× 6c+ 0× 18c = 28c = 28/30 = 14/15

6. X ∼ Poisson(5) and Y = min(X, 12).

(a) P (X = n) = 5ne−5

n!, n = 0, 1, 2, ...

(b) E[X] = λ = 5

(c) V ar(X) = λ = 5

(d)

P (Y = n) =

5ne−5

n!n = 0, 1, 2, ..., 11∑∞

k=12 5ke−5

k!n = 12

0 n = 13, 14...

2

(e)

E[Y ] =

∞∑k=0

kP (Y = k)

=

11∑k=0

k5ke−5

k!+ 12

∞∑k=12

5ke−5

k!

= 5×11∑k=1

5k−1e−5

(k − 1)!+ 12

∞∑k=12

5ke−5

k!

= 12×10∑j=0

5je−5

(j)!+ 12

∞∑k=12

5ke−5

k!− 7×

10∑j=0

5je−5

(j)!

= 12×∞∑k=0

5ke−5

k!− 12× 511

e−5

11!− 7×

10∑j=0

5je−5

(j)!

≈ 12− 12× 0.00824− 6.90413

= 4.99699

7. (a) We know that

1 =

∫ ∞0

ce−5ydy = c/5

so, c = 5 (Y is an exponential random variable with rate 5) .

(b) E[Y ] = 1/5 and V ar(Y ) = 1/25The squared coefficient of variation of Y : Var(Y )/(E[Y ])2) = 1/25/(1/5)2 = 1

(c) Pr{Y > 8} =∫∞8

5e−5ydy = e−40

(d) Pr{Y > 8 | Y > 3} =Pr{Y > 8, Y > 3}

Pr{Y > 3}=Pr{Y > 8}Pr{Y > 3}

=e−40

e−15= e−25

(Pr{Y > 3} is also calculated similar to part (c)).

(e) We know that q satisfies the following0.7 = Pr{Y > q} =

∫∞q

5e−5ydy = e−5q

Then, q = ln(0.7)/(−5) ≈ 0.071

8. fX,Y (x, y) = ce−(3x+4y) for 0 ≤ x and 0 ≤ y.

(a) We know that

1 =

∫ ∞0

∫ ∞0

ce−(3x+4y)dxdy = c/12

Thus, c = 12

(b) Pr{X = Y } = 0 (as both of them are continuous random variables. You can see this if you try toset up the limits of the integration).

(c)

Pr{min(X,Y ) > 1/2} = Pr{X > 1/2, Y > 1/2} =

∫ ∞1/2

∫ ∞1/2

12e−(3x+4y)dxdy = e−1.5e−2 = e−3.5

(d)

Pr{X ≤ Y } =

∫ ∞0

∫ ∞x

12e−3xe−4ydydx =

∫ ∞0

3e−7xdx = 3/7

3

(e) fY (y) =∫∞x

12e−3x−4ydx = 5ne−5

n!= 4e−4y

fY (y) =

{0 y < 0

4e−4y y ≥ 0

E[XY ] =

∫ ∞0

∫ ∞x

xy12e−3xe−4ydydx

=

∫ ∞0

12xe−3x∫ ∞0

ye−4ydy

= 12

∫ ∞0

xe−3x1/16dx

= 12× 1/9× 1/16

= 1/12

9. Let Xk denote the lifetime of the kth bulb where 1 ≤ k ≤ 20.Note that the total lenght of use is

∑20k=1Xk. Then, using the Central Limit Theorem (CTL) and the

standard normal table:

P (

20∑k=1

Xk ≥ 800) = P (

∑20k=1Xk − 20× 30√

20× 30>

800− 20× 30√20× 30

)

≈ P (N(0, 1) > 1.49) ≈ 1− Φ(1.49) ≈ 1− 0.9313 ≈ 0.0687

P (

20∑k=1

Xk ≥ 1000) = P (

∑20k=1Xk − 20× 30√

20× 30>

1000− 20× 30√20× 30

)

≈ P (N(0, 1) > 2.98 = 1− Φ(2.98) ≈ 1− 0.9986 ≈ 0.0014

P (

20∑k=1

Xk ≥ 1200) = P (

∑20k=1Xk − 20× 30√

20× 30>

1200− 20× 30√20× 30

)

≈ P (N(0, 1) > 1.49 = 1− Φ(4.47) ≈ 0

10. X ∼ exp(λ), V ar(X) = E[X2]− (E[X])2

Thus, we will find E[X] and E[X2]E[X] =

∫∞0xe−λxdx

Using integration by parts:

u = xλ v =−e−λx

λdu = λdx dv = e−λxdx

E[X] =[(−xe−λx)

]∞0−∫∞0−e−λxdx = 0− (−1/λ) = 1/λ

E[X2] =∫∞0x2e−λxdx

Using integration by parts:

u = x2λ v =−e−λx

λdu = 2xλdx dv = e−λxdx

E[X2] =[(−x2e−λx)

]∞0− 2/λ

∫∞0−xλe−λxdx = 0 + (2/λ)(1/λ) = 2/λ2

Then, V ar(X) = E[X2]− (E[X])2 = 2/λ2 − 1/λ2 = 1/λ2

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