Linear combinations of random variables Problem set A Worked solutions

The random variable \(X\) has \(E(X)=8\) and \(\mathrm{Var}(X)=2.5\). Find \(E(Y)\) and \(\mathrm{Var}(Y)\) when

1. (a) \(Y=4X+7\)

   • \(E(Y)=4(8)+7=39\)
   • \(\mathrm{Var}(Y)=4^2(2.5)=40\)

2. (b) \(Y=2-5X\)

   • \(E(Y)=-5(8)+2=-38\)
   • \(\mathrm{Var}(Y)=(-5)^2(2.5)=62.5\)

\(X\) and \(Y\) are independent. \(E(X)=5\), \(\mathrm{Var}(X)=2\), \(E(Y)=3\) and \(\mathrm{Var}(Y)=4\). Find \(E(Z)\) and \(\mathrm{Var}(Z)\) when

1. (a) \(Z=2X+3Y-1\)

   • \(E(Z)=2(5)+3(3)-1=18\)
   • \(\mathrm{Var}(Z)=2^2(2)+3^2(4)=44\)

2. (b) \(Z=\dfrac{X-2Y+6}{2}\)

   • \(E(Z)=(5-2(3)+6)/2=2.5\)
   • \(\mathrm{Var}(Z)=(2+(-2)^2(4))/2^2=4.5\)

\(X_1, X_2, \dots\) are independent observations of a random variable \(X\) with \(E(X)=7.5\) and \(\mathrm{Var}(X)=2.2\). Find the mean and variance of

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

   • \(E(S)=4(7.5)=30\)
   • \(\mathrm{Var}(S)=4(2.2)=8.8\)

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

   • \(E(D)=7.5-7.5=0\)
   • \(\mathrm{Var}(D)=2.2+2.2=4.4\)

\(X\), \(Y\) and \(Z\) are independent random variables. \(E(X)=4\), \(\mathrm{Var}(X)=1.5\); \(E(Y)=10\), \(\mathrm{Var}(Y)=2.4\); \(E(Z)=3\), \(\mathrm{Var}(Z)=0.5\). Find the mean and variance of \(W=2X-Y+5Z\).

• \(E(W)=2(4)-10+5(3)=13\)
• \(\mathrm{Var}(W)=2^2(1.5)+2.4+5^2(0.5)=20.9\)

The time to deliver one parcel is a random variable with mean 12.5 minutes and standard deviation 3 minutes. Deliveries are independent. Let \(S\) be the total time for 6 deliveries.

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

   • \(E(S)=6(12.5)=75\)
   • \(\mathrm{Var}(S)=6(3^2)=54\)
   • \(\text{sd}=\sqrt{54}=7.35\)

2. (b) A driver estimates the total by timing one delivery and multiplying by 6. Find the mean and standard deviation of this estimate.

   • \(E(6X)=6(12.5)=75\)
   • \(\mathrm{Var}(6X)=6^2(3^2)=324\)
   • \(\text{sd}=6(3)=18\)

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

   • \(324\div 54=6\)

The mass of a melon has mean 420 g and standard deviation 25 g. The mass of a mango has mean 250 g and standard deviation 18 g. Let \(D\) be the mass of three times one melon minus the total mass of five mangoes, all chosen independently. Find the mean and variance of \(D\), and the standard deviation of \(D\).

• \(E(D)=3(420)-5(250)=10\)
• \(\mathrm{Var}(D)=3^2(625)+5(324)=7245\)
• \(\text{sd}=\sqrt{7245}=85.1\)

The random variable \(X\) has mean 6 and variance 4. The random variable \(Y=aX+b\), with \(a>0\), has mean 40 and variance 36. Find \(a\) and \(b\).

• \(\mathrm{Var}(Y)=a^2(4)=36, \ a^2=9\)
• \(a=3\ (a>0)\)
• \(E(Y)=3(6)+b=40\)
• \(b=40-18=22\)

\(X\) and \(Y\) are independent. \(\mathrm{Var}(X)=5\) and \(\mathrm{Var}(X+Y)=17\). Find \(\mathrm{Var}(Y)\), and then \(\mathrm{Var}(3X-2Y)\).

• \(\mathrm{Var}(X+Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)=17\Rightarrow \mathrm{Var}(Y)=17-5=12\)
• \(\mathrm{Var}(3X-2Y)=3^2(5)+2^2(12)=93\)