1. The system working means that component 1 works, and 1 of components 2, 3, 4 work, and 1 of
components 5, 6 work. We know that X1 ∼ Exp(λ = 2), X2 , X3 , X4 ∼ Exp(λ = 1), and X5 , X6 ∼
Exp(λ = 2/3). So P (X1 < x) = F1 (x) = 1 − exp{−2x} and similarly, F2,3,4 (x) = 1 − exp{−x}, and
F5,6 = 1 − exp{−2/3x}.
(a) P(system lasts ≥ 2 years)
= P(component 1 lasts ≥ 2 years AND at least one of components 2, 3, 4 lasts ≥ 2 years AND
at least one of components 5, 6 lasts ≥ 2 years)
= P (X1 ≥ 2)P (at least one of X2 , X3 , X4 ≥ 2)P (at least one of X5 , X6 ≥ 2)
= P (X1 ≥ 2) (1 − P (all of X2 , X3 , X4 < 2)) (1 − P (both of X5 , X6 < 2))
= (1 − F1 (2)) 1 − F2,3,4 (2)3 1 − F5,6 (2)2
= exp{−2 ∗ 2}(1 − exp{−2}3 )(1 − exp{−2/3 ∗ 2}2 )
= 0.003889
(b) P(1 years ≤ system lasts ≤ 1.25 years)
= P(system lasts ≤ 1.25 years) - P(system lasts ≤ 1 years)
= 1-P(system lasts > 1.25 years) - (1-P(system lasts > 1 years))
= P(system lasts > 1 years)-P(system lasts > 1.25 years)
= exp{−2 ∗ 1}(1 − exp{−1}3 )(1 − exp{−2/3 ∗ 1}2 )
− exp{−2 ∗ 1.25}(1 − exp{−1.25}3 )(1 − exp{−2/3 ∗ 1.25}2 )
= 0.11 − 0.07 = 0.0376
If you calculated P(2 years ≤ system lasts ≤ 2.25 years), your answer should be 0.0022, although
technically this is failing within the first three months of the 3rd year.
2. (a) A linear function of a normally distributed random variable is also normally distributed. This
means that Y is normally distributed, with
µY = EY = E[5X + 2] = 5EX + 2 = 5 ∗ 3 + 2 = 17,
and variance
σY2 = V ar(Y ) = V ar(5X + 2) = 25V ar(X) = 25 ∗ 4 = 100.
So,
Y ∼ N (17, 100)
.
(b)
10 − µY
10 − 17
Y − µY
<
=P Z<
= Φ(−0.7) = 0.242
σY
σY
10
X − µX
10 − µX
10 − 3
P (X < 10) = P
<
=P Z<
= Φ(3.5) = 0.9998
σX
σX
2
P (Y < 10) = P
(c) P (Y < y.99 ) = 0.99 → P z <
(d) P (X < x.99 ) = 0.99 → P z <
y.99 −17
10
x.99 −3
2
= 0.99 →
= 0.99 →
y.99 −17
= 2.33 → y.99 = 2.33 ∗ 10 + 17 = 40.3
10
x.99 −3
= 2.33 → x.99 = 2.33 ∗ 2 + 3 = 7.66
2
(e) W = exp(Y) is log-normal, because log(W) = Y is distributed normal (check your notes!). Therefore,
µW = EW = exp{µY + σY2 /2} = exp{17 + 100/2} = 1.25 ∗ 1029
2
σW
= V ar(W ) = exp{2µY + σY2 }(exp{σY2 } − 1) = 4.22 ∗ 10101
1
3. (a) The easiest way to solve this is by using the multinomial distribution:
n!
× px1 1 . . . pxkk
x1 !, . . . , xk !
12!
P (X1 = 2, X2 = 2, X3 = 2, X4 = 2, X5 = 2, X6 = 2) =
0.242 0.132 0.162 0.202 0.132 0.142 = 0.0132
2!2!2!2!2!2!
P (X1 = x1 , X2 = x2 , . . . , Xk = xk ) =
(b) X = the number of orange candies picked (with and orange being a success, and non-orange being
a failure), so X ∼ Bin(20, .2).
P (X ≤ 5) =
5 X
20
x=0
x
0.20x (1 − 0.20)20−x = 0.804.
(c) X = the number of orange, blue or green picked, so X ∼ Bin(20, 0.2 + 0.24 + 0.16), so
P (X ≥ 10) =
20 X
20
0.60x (1 − 0.60)20−x = 0.872.
x
x=10
4. (a) Yes, covariance ranges from −∞ to ∞.
(b) Yes, if Cov(X, Y ) < 0, then the correlation is negative, since the covariance is divided by the
standard deviations, which are always positive.
(c) For the same reason the previous problem is true, this is also true.
(d) No, correlation always ranges between -1 and 1.
(e) See above.
(f) Cov(aX, bY) = abCov(X,Y). Proof:
Cov(aX, bY ) = E(aX ∗ bY ) − E(aX)E(bY )
= abE(XY ) − abE(X)E(Y )
= ab (E(XY ) − EX ∗ EY )
= abCov(X, Y )
So, if Cov(X,Y) = 0.3, then Cov(100X, Y) = 100Cov(X,Y) = 100*0.3 = 30.
(g) Corr(aX, bY) = sgn(a*b)corr(X,Y). Proof:
Corr(aX, bY ) = p
=p
Cov(aX, bY )
V ar(aX)V ar(bY )
abCov(X, Y )
a2 b2 V ar(X)V ar(Y )
abCov(X, Y )
= p
ab V ar(X)V ar(Y )
= sgn(ab)Cov(X, Y )
Therefore, Corr(100X, Y) = +Corr(X, Y) = 0.1.
(h) Cov(X, X) = Var(X). Proof:
Cov(X, X) = E(X ∗ X) − E(X)E(X) = E(X 2 ) − (EX)2 = V ar(X).
(i) Corr(X, X) = 1. Proof:
Corr(X, X) = p
Cov(X, X)
V ar(X)V ar(X)
2
V ar(X)
=p
= 1.
V ar(X)2
(j) Cov(100X, 10X) = 100*10*var(X) = 1000Var(X).
(k) Corr(100X, 10X) = sgn(100*10)*(1) = 1.
(l) The correlation measures the linear relationship between two variables. There is no linear relationship between X and X 2 , therefore the correlation is 0.
5. (a) EX1 = E(w + E1 ) = w + E(E1 ) = w + 0 = w. Similarly, EX2 = w.
(b) V ar(X1 ) = V ar(w + E1 ) = V ar(E1 ) = 0.1. Similarly, V ar(X2 ) = 0.1.
(c) The correlation between the two variables is 0, since they are independent (because E1 and E2
are independent). Proof:
Cov(X1 , X2 )
Corr(X1 , X2 ) = p
V ar(X1 )V ar(X2 )
E(X1 X2 ) − EX1 ∗ EX2
=
0.1
E ((w + E1 )(w + E2 )) − w2
=
0.1
E w2 + wE1 + wE2 + E1 ∗ E2 − w2
=
0.1
w2 + 0 + 0 + E(E1 ∗ E2 ) − w2
=
0.1
w2 + E(E1 )E(E2 ) − w2
*If two r.v.s X and Y are independent, then E(XY ) = EX ∗ EY
=
0.1
2
2
w +0∗0−w
=
=0
0.1
(d) We can generate normally distributed data using the rnorm() function in R. We also need to pick
a hypothetical value for w, in this case I will pick w = 5. I will first generate 10 samples of X1
and X2 :
> w = 5
> samplesize = 10
> X1 = w + rnorm(samplesize, mean = 0, sd = sqrt(0.1))
> X2 = w + rnorm(samplesize, mean = 0, sd = sqrt(0.1))
> mean(X1)
[1] 4.997936
> mean(X2)
[1] 5.000257
> var(X1)
[1] 0.03333543
> var(X2)
[1] 0.1490419
> cor(X1, X2)
[1] -0.2196698
The easiest way to examine how this changes based on sample size is through plotting different
values. I will calculate and store the mean, variance, and correlation values in a loop first, and
then plot them:
>
>
>
+
+
+
#Make a place to store all the values:
keepmeanX1 = keepmeanX2 = keepvarX1 = keepvarX2 = keepcor = NULL
for(i in 1:1000){
# The sample size increases by 10 each time.
X1 = w + rnorm(10*i, mean = 0, sd = sqrt(0.1))
X2 = w + rnorm(10*i, mean = 0, sd = sqrt(0.1))
3
+ # Store the values:
+ keepmeanX1 = c(keepmeanX1, mean(X1))
+ keepmeanX2 = c(keepmeanX2, mean(X2))
+ keepvarX1 = c(keepvarX1, var(X1))
+ keepvarX2 = c(keepvarX2, var(X2))
+ keepcor = c(keepcor, cor(X1, X2))
+ }
>
> par(mfrow = c(1,3)) # Put the mean, variance, and correlation plots on one graph.
> plot(1:1000*samplesize, keepvarX1, type = ’l’,
main = ’Variance’, xlab = ’sample size’, ylab = ’Var’)
> abline(h = 0.1, col = ’red’)
> plot(1:1000*samplesize, keepmeanX1, type = ’l’,
main = ’Means’, xlab = ’sample size’, ylab = ’Mean’)
> abline(h = 5, col = ’red’)
> plot(1:1000*samplesize, keepvarX1, type = ’l’,
main = ’Variance’, xlab = ’sample size’, ylab = ’Var’)
> abline(h = 0.1, col = ’red’)
> plot(1:1000*samplesize, keepcor, type = ’l’,
main = ’Correlation’, xlab = ’sample size’, ylab = ’Corr’)
> abline(h = 0, col = ’red’)
Variance
Correlation
0.10
-0.10
0.09
2000
4000
6000
sample size
8000
10000
-0.20
0.08
0
0.00
Corr
0.11
Var
0.10
5.00
4.95
4.90
Mean
5.05
0.12
5.10
Means
0
2000
4000
6000
sample size
8000
10000
0
2000
4000
6000
8000
10000
sample size
In the figure, we can see that as the sample size increases, the values of the mean, variance, and
correlation approach the true theoretical values we derived in the previous parts of this problem.
4