Linear combinations of random variables Problem set B Worked solutions

The random variable \(X\) has \(E(X)=6.4\) and \(\mathrm{Var}(X)=1.6\). Find \(E(Y)\) and \(\mathrm{Var}(Y)\) when

1. (a) \(Y=6X-5\)

   • \(E(Y)=6(6.4)-5=33.4\)
   • \(\mathrm{Var}(Y)=6^2(1.6)=57.6\)

2. (b) \(Y=9-3X\)

   • \(E(Y)=-3(6.4)+9=-10.2\)
   • \(\mathrm{Var}(Y)=(-3)^2(1.6)=14.4\)

\(X\) and \(Y\) are independent. \(E(X)=12\), \(\mathrm{Var}(X)=5\), \(E(Y)=7\) and \(\mathrm{Var}(Y)=3\). Find \(E(Z)\) and \(\mathrm{Var}(Z)\) when

1. (a) \(Z=4X-Y-3\)

   • \(E(Z)=4(12)-7-3=38\)
   • \(\mathrm{Var}(Z)=4^2(5)+3=83\)

2. (b) \(Z=\dfrac{3X+Y-10}{4}\)

   • \(E(Z)=(3(12)+7-10)/4=8.25\)
   • \(\mathrm{Var}(Z)=(3^2(5)+3)/4^2=3\)

\(X_1, X_2, \dots\) are independent observations of a random variable \(X\) with \(E(X)=0.8\) and \(\mathrm{Var}(X)=0.35\). Find the mean and variance of

1. (a) \(S=X_1+X_2+X_3+X_4+X_5\)

   • \(E(S)=5(0.8)=4\)
   • \(\mathrm{Var}(S)=5(0.35)=1.75\)

2. (b) \(D=X_1-X_2\)

   • \(E(D)=0.8-0.8=0\)
   • \(\mathrm{Var}(D)=0.35+0.35=0.7\)

\(X\), \(Y\) and \(Z\) are independent random variables. \(E(X)=9\), \(\mathrm{Var}(X)=0.8\); \(E(Y)=2\), \(\mathrm{Var}(Y)=0.3\); \(E(Z)=6\), \(\mathrm{Var}(Z)=1.9\). Find the mean and variance of \(W=3X+4Y-Z\).

• \(E(W)=3(9)+4(2)-6=29\)
• \(\mathrm{Var}(W)=3^2(0.8)+4^2(0.3)+1.9=13.9\)

The mass of a loaf from a bakery is a random variable with mean 640 g and standard deviation 15 g. Loaves are independent. Let \(S\) be the total mass of 4 loaves.

1. (a) Find the mean and standard deviation of \(S\).

   • \(E(S)=4(640)=2560\)
   • \(\mathrm{Var}(S)=4(15^2)=900\)
   • \(\text{sd}=\sqrt{900}=30\)

2. (b) A shopper estimates the total by weighing one loaf and multiplying by 4. Find the mean and standard deviation of this estimate.

   • \(E(4X)=4(640)=2560\)
   • \(\mathrm{Var}(4X)=4^2(15^2)=3600\)
   • \(\text{sd}=4(15)=60\)

3. (c) Find the variance of the estimate in part (b) divided by the variance of \(S\).

   • \(3600\div 900=4\)

A runner's time for one lap of a track has mean 95 seconds and standard deviation 4 seconds. A cyclist's time for one lap has mean 38 seconds and standard deviation 2.5 seconds. Let \(D\) be twice one lap time of the runner minus the total time of six laps of the cyclist, all independent. Find the mean and variance of \(D\), and the standard deviation of \(D\).

• \(E(D)=2(95)-6(38)=-38\)
• \(\mathrm{Var}(D)=2^2(16)+6(6.25)=101.5\)
• \(\text{sd}=\sqrt{101.5}=10.1\)

The random variable \(X\) has mean 15 and variance 2.5. The random variable \(Y=aX+b\), with \(a>0\), has mean 53 and variance 40. Find \(a\) and \(b\).

• \(\mathrm{Var}(Y)=a^2(2.5)=40, \ a^2=16\)
• \(a=4\ (a>0)\)
• \(E(Y)=4(15)+b=53\)
• \(b=53-60=-7\)

\(P\) and \(Q\) are independent. \(\mathrm{Var}(P)=6\) and \(\mathrm{Var}(P+Q)=20\). Find \(\mathrm{Var}(Q)\), and then \(\mathrm{Var}(2P-5Q)\).

• \(\mathrm{Var}(P+Q)=\mathrm{Var}(P)+\mathrm{Var}(Q)=20\Rightarrow \mathrm{Var}(Q)=20-6=14\)
• \(\mathrm{Var}(2P-5Q)=2^2(6)+5^2(14)=374\)