Wize AP Statistics Textbook > Random Variables and Probability Distributions

Combining Random Variables

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Linear Combinations of Random Variables
For a random variable XXX, let...
E(X)=E\left(X\right)=E(X)= expected value or mean
Var(X)=Var\left(X\right)=Var(X)= variance
SD(X)=SD\left(X\right)=SD(X)= standard deviation

A new distribution could be formed by combining random variables Given the mean and standard deviation a given distribution, we can combine random variables to find the combined mean, combined variance, and combined standard deviation of the new distribution.
Linear Rescaling of RV 
A linear rescaling is a transformation of the form Y=aX+bY=a\textcolor{orange}{X}+bY=aX+b

E(aX+b)=aE(X)+bE\left(a\textcolor{orange}{X}+b\right)=aE\left(\textcolor{orange}{X}\right)+bE(aX+b)=aE(X)+b
Var(aX+b)=a2Var(X)Var\left(a\textcolor{orange}{X}+b\right)=a^2Var\left(\textcolor{orange}{X}\right)Var(aX+b)=a2Var(X)
SD(aX+b)=∣a∣Var(X)SD(a\textcolor{orange}{X}+b)=|a|\sqrt{Var(\textcolor{orange}{X})}SD(aX+b)=∣a∣Var(X)​


Combining RV for Weighted Averages

A linear combination of two random variables XXX and YYY is in the form aX+bYa\textcolor{orange}{X}+b\textcolor{green}{Y}aX+bY where a weighted average is involved such that a+b=1a+b=1a+b=1.
E(aX+bY)=aE(X)+bE(Y)E\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=aE\left(\textcolor{orange}{X}\right)+bE\left(\textcolor{green}{Y}\right)E(aX+bY)=aE(X)+bE(Y)
Var(aX+bY)=a2Var(X)+b2Var(Y)Var\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)Var(aX+bY)=a2Var(X)+b2Var(Y)
SD(aX+bY)=a2Var(X)+b2Var(Y)SD\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=\sqrt{a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)}SD(aX+bY)=a2Var(X)+b2Var(Y)​

E(aX−bY)=aE(X)−bE(Y)E\left(a\textcolor{orange}{X}-b\textcolor{green}{Y}\right)=aE\left(\textcolor{orange}{X}\right)-bE\left(\textcolor{green}{Y}\right)E(aX−bY)=aE(X)−bE(Y)
Var(aX−bY)=a2Var(X)+b2Var(Y)Var\left(a\textcolor{orange}{X}-b\textcolor{green}{Y}\right)=a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)Var(aX−bY)=a2Var(X)+b2Var(Y)
SD(aX−bY)=a2Var(X)+b2Var(Y)SD\left(a\textcolor{orange}{X}-b\textcolor{green}{Y}\right)=\sqrt{a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)}SD(aX−bY)=a2Var(X)+b2Var(Y)​


Wize Tip
The standard deviation is the square-root of the variance.

When combining random variables, always add variances. You cannot subtract variances.

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Example: Linear Combination of Random Variables

Let LLL be a discrete random variable where 𝐸(𝐿)=2,SD(𝐿)=5.𝐸(𝐿)=2,SD(𝐿)=5.E(L)=2,SD(L)=5. 
E(aX+b)=aE(X)+bE\left(a\textcolor{orange}X+b\right)=aE\left(\textcolor{orange}X\right)+bE(aX+b)=aE(X)+b
Var(aX+b)=a2Var(X)Var\left(a\textcolor{orange}{X}+b\right)=a^2Var\left(\textcolor{orange}{X}\right)Var(aX+b)=a2Var(X)
SD(aX+b)=∣a∣Var(X)SD(a\textcolor{orange}{X}+b)=|a|\sqrt{Var(\textcolor{orange}{X})}SD(aX+b)=∣a∣Var(X)​

(a) Find 𝐸(3𝐿−4)𝐸(3𝐿 − 4)E(3L−4) 

E(aX+b)=aE(X)+bE\left(a\textcolor{orange}X+b\right)=aE\left(\textcolor{orange}X\right)+bE(aX+b)=aE(X)+b

𝐸(3𝐿−4)=3(2)−4=2𝐸(3𝐿 − 4) = 3(2) − 4 = 2E(3L−4)=3(2)−4=2

(b) Find SD(−𝐿+12)SD(−𝐿+12)SD(−L+12)

SD(aX+b)=∣a∣V(X)SD(a\textcolor{orange}{X}+b)=|a|\sqrt{V(\textcolor{orange}{X})}SD(aX+b)=∣a∣V(X)​

Tip: Imagine an invisible 1 in front of the LLL, so a=−1a=-1a=−1.

SD(−𝐿+12)=∣−1∣SD(𝐿)=(1)(5)=5SD(−𝐿+12)=|−1|SD(𝐿)=(1)(5)=5SD(−L+12)=∣−1∣SD(L)=(1)(5)=5

I personally prefer to solve for the variance first, then find the square-root of the variance to solve for the standard deviation:

V(aX+b)=a2V(X)V\left(a\textcolor{orange}{X}+b\right)=a^2V\left(\textcolor{orange}{X}\right)V(aX+b)=a2V(X)

Var(−𝐿+12)=(−1)2(5)2=25Var(−𝐿+12)=\left(-1\right)^2\left(5\right)^2=25Var(−L+12)=(−1)2(5)2=25
SD=Var=25=5SD=\sqrt{Var}=\sqrt{25}=5SD=Var​=25​=5

Let QQQ be a discrete random variable where 𝐸(Q)=25,SD(Q)=6.𝐸(Q)=25,SD(Q)=6.E(Q)=25,SD(Q)=6. 

(a) Find 𝐸(Q2−1)𝐸\left(\frac{Q}{2}−1\right)E(2Q​−1) 

(b) Find SD(Q2−1)SD\left(\frac{Q}{2}−1\right)SD(2Q​−1)

(a)

(b)

I don't know

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Example: Combining Random Variables with Weighted Averages

Conrad invests in two portfolios, 40% in the Aggressive Portfolio and 60% in the Slow-and-Steady Portfolio. His wealth manager determined that the average rate of return in the Aggressive Portfolio is 14.5% with a standard deviation of 20% – risky! The Slow-and-Steady Portfolio has an average rate of return of 3.3% with a standard deviation of 1%. 

Determine the mean rate of return, variance, and standard deviation of his portfolios. (Assume the two portfolios are independent.)
E(aX+bY)=aE(X)+bE(Y)E\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=aE\left(\textcolor{orange}{X}\right)+bE\left(\textcolor{green}{Y}\right)E(aX+bY)=aE(X)+bE(Y)
Var(aX+bY)=a2Var(X)+b2Var(Y)Var\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)Var(aX+bY)=a2Var(X)+b2Var(Y)
SD(aX+bY)=a2Var(X)+b2Var(Y)SD\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=\sqrt{a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)}SD(aX+bY)=a2Var(X)+b2Var(Y)​

a=0.4 a=0.4\ a=0.4 
b=0.6b=0.6b=0.6

Let Portfolio P=0.4X+0.6YP=0.4X+0.6YP=0.4X+0.6Y

Combined mean:

E(aX+bY)=aE(X)+bE(Y)E\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=aE\left(\textcolor{orange}{X}\right)+bE\left(\textcolor{green}{Y}\right)E(aX+bY)=aE(X)+bE(Y)

E(P)=0.4(14.5)+0.6(3.3)=7.78E\left(P\right)=0.4\left(14.5\right)+0.6\left(3.3\right)=7.78E(P)=0.4(14.5)+0.6(3.3)=7.78 (or 7.78%)

Combined variance:

Var(aX+bY)=a2Var(X)+b2Var(Y)Var\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)Var(aX+bY)=a2Var(X)+b2Var(Y)

Var(P)=(0.4)2(20)2+(0.6)2(1)2=64.36Var\left(P\right)=\left(0.4\right)^2\left(20\right)^2+\left(0.6\right)^2\left(1\right)^2=64.36Var(P)=(0.4)2(20)2+(0.6)2(1)2=64.36 (or 64.36%)

Combined standard deviation:

SD(aX+bY)=a2Var(X)+b2Var(Y)SD\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=\sqrt{a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)}SD(aX+bY)=a2Var(X)+b2Var(Y)​

SD(P)=Var(P)=64.36=8.022SD\left(P\right)=\sqrt{Var\left(P\right)}=\sqrt{64.36}=8.022SD(P)=Var(P)​=64.36​=8.022 (or 8.022%)

PAGE BREAK

Alternative Method:

Combined mean:

E(aX+bY)=aE(X)+bE(Y)E\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=aE\left(\textcolor{orange}{X}\right)+bE\left(\textcolor{green}{Y}\right)E(aX+bY)=aE(X)+bE(Y)

E(P)=0.4(0.145)+0.6(0.033)=0.0778E\left(P\right)=0.4\left(0.145\right)+0.6\left(0.033\right)=0.0778E(P)=0.4(0.145)+0.6(0.033)=0.0778 (or 7.78%)

Combined variance:

Var(aX+bY)=a2Var(X)+b2Var(Y)Var\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)Var(aX+bY)=a2Var(X)+b2Var(Y)

Var(P)=(0.4)2(0.2)2+(0.6)2(0.01)2=0.006436Var\left(P\right)=\left(0.4\right)^2\left(0.2\right)^2+\left(0.6\right)^2\left(0.01\right)^2=0.006436Var(P)=(0.4)2(0.2)2+(0.6)2(0.01)2=0.006436* 

*Note: The variance’s usefulness is limited here because the units are squared and not the same as the original data.

Combined standard deviation:

SD(aX+bY)=a2Var(X)+b2Var(Y)SD\left(a\textcolor{orange}{X}+b\textcolor{green}{Y}\right)=\sqrt{a^2Var\left(\textcolor{orange}{X}\right)+b^2Var\left(\textcolor{green}{Y}\right)}SD(aX+bY)=a2Var(X)+b2Var(Y)​

SD(P)=Var(P)=0.006436=0.08022SD\left(P\right)=\sqrt{Var\left(P\right)}=\sqrt{0.006436}=0.08022SD(P)=Var(P)​=0.006436​=0.08022 (or 8.022%)

Practice: Combining Random Variables with Weighted Average
The midterm is worth 30% of your overall grade; the final exam is worth 70%. 

The average grade on the midterm is 68% with a standard deviation of 12%.
The average grade on the final exam is 73% with a standard deviation of 16%

(a) What is the average overall grade? [Provide answer with 2 decimal places, e.g. 65.2]

(b) What is the standard deviation of the overall grade? [Provide answer with 2 decimal places, e.g. 65.2]

I don't know

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Functions of Two Independent Random Variables X and Y
Two random variables XXX and YYY are independent if knowing the value of one of them does not affect the probabilities for the other one.
If X and Y are independent, we have these formulas:
E(XY)=E(X)×E(Y)E(\textcolor{orange}{X}\textcolor{blue}{Y})=\textcolor{orange}{E(X)}\times\textcolor{blue}{E(Y)}E(XY)=E(X)×E(Y)

E(X+Y)=E(X)+E(Y)E(\textcolor{orange}{X}\textcolor{blue}{+Y})=\textcolor{orange}{E(X)}+\textcolor{blue}{E(Y)}E(X+Y)=E(X)+E(Y) 
E(X−Y)=E(X)−E(Y)E(\textcolor{orange}{X}\textcolor{blue}{-Y})=\textcolor{orange}{E(X)}-\textcolor{blue}{E(Y)}E(X−Y)=E(X)−E(Y) 

Var(X+Y)=Var(X)+Var(Y)Var(\textcolor{orange}{X}+\textcolor{blue}{Y})=\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}Var(X+Y)=Var(X)+Var(Y)
Var(X−Y)=Var(X)+Var(Y)Var(\textcolor{orange}{X}-\textcolor{blue}{Y})=\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}Var(X−Y)=Var(X)+Var(Y) 

Watch Out!
V(X−Y)=V(X)+V(Y)V(\textcolor{orange}{X}-\textcolor{blue}{Y})=\textcolor{orange}{V(X)}+\textcolor{blue}{V(Y)}V(X−Y)=V(X)+V(Y)
This formula is NOT a mistake! Never subtract variances! Always add variances.




The standard deviation is the square-root of the variance:
SD(X+Y)=Var(X)+Var(Y)SD(\textcolor{orange}{X}+\textcolor{blue}{Y})=\sqrt{\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}}SD(X+Y)=Var(X)+Var(Y)​
SD(X−Y)=Var(X)+Var(Y)SD(\textcolor{orange}{X}-\textcolor{blue}{Y})=\sqrt{\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}}SD(X−Y)=Var(X)+Var(Y)​ 

Watch Out!
Do not add or subtract standard deviations! Always combine the variances first (by adding) and then find the square-root of the combined variances to solve for the combined standard deviation.

As you can see, there are a lot of formulas but they will all make sense after we do a bunch of examples.

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Example: Combining Independent Random Variables
Let 𝐴𝐴A, 𝐵𝐵B be two discrete independent random variables with:

 E(A)=10, SD(𝐴)=5E\left(A\right)=10,\ SD(𝐴)=5E(A)=10, SD(A)=5 
 E(B)=26, SD(𝐵)=12E\left(B\right)=26,\ SD(𝐵)=12E(B)=26, SD(B)=12

E(XY)=E(X)×E(Y)E(\textcolor{orange}X\textcolor{blue}Y)=\textcolor{orange}{E(X)}\times \textcolor{blue}{E(Y)}E(XY)=E(X)×E(Y)

E(X+Y)=E(X)+E(Y)E(\textcolor{orange}{X}\textcolor{blue}{+Y})=\textcolor{orange}{E(X)}+\textcolor{blue}{E(Y)}E(X+Y)=E(X)+E(Y) 
E(X−Y)=E(X)−E(Y)E(\textcolor{orange}{X}\textcolor{blue}{-Y})=\textcolor{orange}{E(X)}-\textcolor{blue}{E(Y)}E(X−Y)=E(X)−E(Y) 

Var(X+Y)=Var(X)+Var(Y)Var(\textcolor{orange}{X}+\textcolor{blue}{Y})=\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}Var(X+Y)=Var(X)+Var(Y)
Var(X−Y)=Var(X)+Var(Y)Var(\textcolor{orange}{X}-\textcolor{blue}{Y})=\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}Var(X−Y)=Var(X)+Var(Y) 
The standard deviation is the square-root of the variance:
SD(X+Y)=SD(X−Y)=Var(X)+Var(Y)SD(\textcolor{orange}{X}+\textcolor{blue}{Y})=SD(\textcolor{orange}{X}-\textcolor{blue}{Y})=\sqrt{\textcolor{orange}{Var(X)}+\textcolor{blue}{Var(Y)}}SD(X+Y)=SD(X−Y)=Var(X)+Var(Y)​

(a) Find E(A+B)E\left(A+B\right)E(A+B)

E(A+B)=E(A)+E(B)=10+26=36E(\textcolor{orange}{A}\textcolor{blue}{+B})=\textcolor{orange}{E(A)}+\textcolor{blue}{E(B)}=10+26=36E(A+B)=E(A)+E(B)=10+26=36

(b) Find SD(A+B)SD(A+B)SD(A+B)

Do not add standard deviations! Add variances first, then find the square-root of the combined variances.

SD(A+B)SD(A+B)SD(A+B)
=Var(A+B)=\sqrt{Var(A+B)}=Var(A+B)​
=Var(A)+Var(B)=\sqrt{Var(A)+Var(B)}=Var(A)+Var(B)​
=52+122=\sqrt{5^2+12^2}=52+122​
=169=\sqrt{169}=169​
=13=13=13

(c) Find SD(2𝐴−𝐵)SD(2𝐴−𝐵)SD(2A−B)

Do not subtract standard deviations! Add variances first, then find the square-root of the combined variances.

SD(2A−B)SD(2A-B)SD(2A−B)
=Var(2A−B)=\sqrt{Var(2A-B)}=Var(2A−B)​
=22Var(A)+(−1)2Var(B)=\sqrt{2^2Var(A)+(-1)^2Var(B)}=22Var(A)+(−1)2Var(B)​
=4Var(A)+1Var(B)=\sqrt{4Var\left(A\right)+1Var\left(B\right)}=4Var(A)+1Var(B)​
=4⋅52+122=\sqrt{4\cdot5^2+12^2}=4⋅52+122​
=100+144=\sqrt{100+144}=100+144​
=244=\sqrt{244}=244​

Suppose that certain car parts are manufactured so that their heights are independent of each other, with a mean of μ=31.2\mu=31.2μ=31.2 cm and a standard deviation of σ=0.1\sigma=0.1σ=0.1 cm.

If 2 of these parts are stacked on top of each other to form a new part, what are the mean and standard deviation of this new part?

Mean = 62.4 cm; Standard deviation = 0.141 cm

Mean = 62.4 cm; Standard deviation = 0.2 cm

Mean = 62.4 cm; Standard deviation = 0.45 cm

Mean = 31.2cm; Standard deviation = 0.02 cm

Mean =  31.2 cm; Standard deviation = 0.15 cm

I don't know

You have 50 job candidates to interview for a job at your firm. Each interview lasts an average of 12 minutes with a standard deviation of 3 minutes. What is the mean and standard deviation of the total time to interview all 50 candidates (not counting breaks in between)?

Combined mean (50 candidates)

Combined standard deviation (50 candidates)

I don't know