Wize AP Statistics Textbook > Random Variables and Probability Distributions

Joint Distribution

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Conditional Probability & Joint Distribution
The joint distribution table of XXX and YYY can be represented using a table:

       (Y=y1)              (Y=y2)            ...     (X=x1)(X=x2)(X=x3)⋮\begin{array}{c|c|c|}
\\&\ \ \ \ \ \ \ (Y=y_1)\ \ \ \ \ \ \ &\ \ \ \ \ \ \ (Y=y_2)\ \ \ \ \ \ \ &\ \ \ \ \ ...\ \ \ \ \ \\\\
\hline\\
(X=x_1)\\\\
\hline\\
(X=x_2)\\\\
\hline\\
(X=x_3)\\\\
\hline
\vdots
\end{array}(X=x1​)(X=x2​)(X=x3​)⋮​       (Y=y1​)              (Y=y2​)            ...     ​

The sum of all the probabilities is 1.


The random variables XXX and YYY are independent if P(X∩Y)=P(X)×P(Y)P(X\cap Y)=P(X)\times P(Y)P(X∩Y)=P(X)×P(Y).

For joint distributions, the random variablesXXX and YYY are independent if P(xi∩yi)=P(xi)×P(yi)P(x_i\cap y_i)=P(x_i)\times P(y_i)P(xi​∩yi​)=P(xi​)×P(yi​) where i=1,2,3..i=1,2,3..i=1,2,3..

Watch Out!
XXX and YYY are independent only if this multiplication rule works in all of the cells in the joint distribution table! If not, then XXX and YYY are not independent. 

Wize Tip
If XXX and YYY are independent random variables, then aXaXaX and bYbYbY are also independent.

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Example: Conditional Probability Formula

You can solve for conditional probabilities when dealing with joint distributions.


P(A∣B)=P(A∩B)P(B)\boxed{\displaystyle P\left(A\mid B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}}P(A∣B)=P(B)P(A∩B)​​

Example:


Given that the opponent plays "water", what is the probability that your Pokey Mon plays "fire"?

P(Fire∣Water)=P(Fire∩Water)P(Water)\displaystyle{P\left(Fire\mid Water\right)=\frac{P\left(Fire\cap Water\right)}{P\left(Water\right)}}P(Fire∣Water)=P(Water)P(Fire∩Water)​

=0.150.15+0.10+0.10=0.150.35=0.4286\displaystyle{=\frac{0.15}{0.15+0.10+0.10}=\frac{0.15}{0.35}=0.4286}=0.15+0.10+0.100.15​=0.350.15​=0.4286

Are the two random variables independent?

The random variables XXX and YYY are independent if P(X∩Y)=P(X)×P(Y)P(X\cap Y)=P(X)\times P(Y)P(X∩Y)=P(X)×P(Y).

P(Fire∩Water)=0.15P\left(Fire\cap Water\right)=0.15P(Fire∩Water)=0.15
P(Fire)×P(Water)=(0.45)(0.35)=0.1575P\left(Fire\right)\times P\left(Water\right)=\left(0.45\right)\left(0.35\right)=0.1575P(Fire)×P(Water)=(0.45)(0.35)=0.1575

0.15≠0.15750.15\ne0.15750.15=0.1575

Therefore, they are not independent. 

Wize Tip
See: Conditional Probability

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Example: Conditional Probability & Joint Distribution

XXX and YYY are random variables and you are working in a busy office on a given day.

X=X=X= number of times your phone rings
Y=Y=Y= number of times someone walks into your office
(i) What is the probability your phone rings 2 times and 1 person walks into your office?

P(X=2∩Y=1)=0.09P\left(X=2\cap  Y=1\right)=0.09P(X=2∩Y=1)=0.09

(ii) Find P(X=0∩Y≥1)P\left(X=0\cap Y\ge1\right)P(X=0∩Y≥1)

P(X=0∩Y≥1)=Px,y(0,1)+Px,y(0,2)=0.15+0.06=0.21P\left(X=0\cap Y\ge1\right)=P_{x,y}\left(0,1\right)+P_{x,y}\left(0,2\right)=0.15+0.06=0.21P(X=0∩Y≥1)=Px,y​(0,1)+Px,y​(0,2)=0.15+0.06=0.21

(iii) Find P(Y=1∣X=0)P\left(Y=1\mid X=0\right)P(Y=1∣X=0) 

Conditional probability formula:

P(A∣B)=P(A∩B)P(B)\displaystyle P\left(A\mid B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}P(A∣B)=P(B)P(A∩B)​

P(Y=1∣X=0)=P(X=0∩Y=1)P(X=0)\displaystyle P\left(Y=1\mid X=0\right)=\frac{P\left(X=0\cap Y=1\right)}{P\left(X=0\right)}P(Y=1∣X=0)=P(X=0)P(X=0∩Y=1)​

=0.060.08+0.15+0.06=0.150.29=0.5172\displaystyle =\frac{0.06}{0.08+0.15+0.06}=\frac{0.15}{0.29}=0.5172=0.08+0.15+0.060.06​=0.290.15​=0.5172

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(iv) Are XXX and YYY independent?

The random variables XXX and YYY are independent if P(xi∩yi)=P(xi)×P(yi)P(x_i\cap y_i)=P(x_i)\times P(y_i)P(xi​∩yi​)=P(xi​)×P(yi​) where i=1,2,3..i=1,2,3..i=1,2,3... (This must be true for all cells.)

Let's test it for X=0,Y=2X=0, Y=2X=0,Y=2:

P(X=0∩Y=2)=0.06P(X=0\cap Y=2)=0.06P(X=0∩Y=2)=0.06

P(X=0)=0.08+0.15+0.06=0.29P(X=0)=0.08+0.15+0.06=0.29P(X=0)=0.08+0.15+0.06=0.29
P(Y=2)=0.06+0.03+0.13+0.04=0.26P(Y=2)=0.06+0.03+0.13+0.04=0.26P(Y=2)=0.06+0.03+0.13+0.04=0.26

(0.29)(0.26)=0.07540.06≠0.0754(0.29)(0.26)=0.0754
\\0.06\neq0.0754(0.29)(0.26)=0.07540.06=0.0754
Therefore, XXX and YYY are not independent.

Practice: Conditional Probability & Joint Distribution

(a)  Find P(X=0∩Y≥0)P\left(X=0\cap Y\ge0\right)P(X=0∩Y≥0)

(b) Find P(Y=0∣X=1)P\left(Y=0\mid X=1\right)P(Y=0∣X=1) 

(c) Are XXX and YYY independent?

(a) Find (X=0, Y>=0). Provide answer with 2 decimal places (e.g. 0.25)

(b) Find (Y=0 | X=1). Provide answer with 2 decimal places (e.g. 0.25)

I don't know

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Random Variables & Joint Distribution
It is a very good idea to review the following formulas as they are useful when dealing with joint probability distributions. 

Review: Expected Value, Variance, and Standard Deviation of a Random Variable X

Let XXX be a discrete random variable.

The expected value or mean of XXX is denoted by E(X)E(X)E(X) or μ\mu μ, and is calculated by
E(X)=μ=∑[xiP(xi)]\boxed{E(X)=\mu=\sum
\left[x_iP(x_i)
\right]}E(X)=μ=∑[xi​P(xi​)]​

The variance of XXX is denoted by Var(X)Var(X)Var(X) or σ2\sigma^2σ2, and is calculated by
Var(X)=σ2=∑[(xi−μ)2P(xi)]\boxed{

Var(X)=\sigma^2=\sum\left[(x_i-\mu)^2P(x_i)\right]}Var(X)=σ2=∑[(xi​−μ)2P(xi​)]​
or
Var(X)=σ2=[∑xi 2P(xi)]−μ2\boxed{
\begin{array}{c}


Var(X)=\sigma^2 =\left[\sum x_i^{\ 2}P(x_i)\right]-\mu^2
\end{array}}Var(X)=σ2=[∑xi 2​P(xi​)]−μ2​​
or
Var(X)=E(X2)−E(X)2\boxed{Var\left(X\right)=E\left(X^2\right)-E\left(X\right)^2}

Var(X)=E(X2)−E(X)2​
where

E(X2)=∑(xi)2P(xi)E(X^2)=\sum(x_i)^2P(x_i)E(X2)=∑(xi​)2P(xi​)

The standard deviation of XXX is denoted by SD(X)SD(X)SD(X) or σ\sigmaσ, and is calculated by
SD(X)=σ=Var(X)\boxed{SD(X)=\sigma=\sqrt{Var(X)}}SD(X)=σ=Var(X)​​

Wize Tip
See: Discrete Random Variables

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Example with one random variable X:
xP(X)10.520.5\begin{array}{|c|c|}

\hline
x&P(X)\\
\hline
1&0.5\\
\hline
2&0.5\\
\hline
\end{array}x12​P(X)0.50.5​​
Mean:

E(X)=μ=∑[xiP(xi)]E(X)=\mu=\sum[x_iP(x_i)]E(X)=μ=∑[xi​P(xi​)]

E(X)=(1)(0.5)+(2)(0.5)=1.5E\left(X\right)=\left(1\right)\left(0.5\right)+\left(2\right)\left(0.5\right)=1.5E(X)=(1)(0.5)+(2)(0.5)=1.5

Variance (Method 1):

Var(X)=σ2=∑(xi−μ)2P(xi)Var(X)=\sigma^2=\sum(x_i-\mu)^2P(x_i)Var(X)=σ2=∑(xi​−μ)2P(xi​)

(1−1.5)2(0.5)+(2−1.5)2(0.5)=0.125+0.125=0.25\left(1-1.5\right)^2\left(0.5\right)+\left(2-1.5\right)^2\left(0.5\right)=0.125+0.125=0.25(1−1.5)2(0.5)+(2−1.5)2(0.5)=0.125+0.125=0.25

Variance (Method 2):

Var(X)=σ2=[∑xi 2P(xi)]−μ2Var(X)=\sigma^2 =\left[\sum x_i^{\ 2}P(x_i)\right]-\mu^2Var(X)=σ2=[∑xi 2​P(xi​)]−μ2

[(1)2(0.5)+(2)2(0.5)]−(1.5)2=[0.5+2]−2.25=2.5−2.25=0.25\left[\left(1\right)^2\left(0.5\right)+\left(2\right)^2\left(0.5\right)\right]-\left(1.5\right)^2=\left[0.5+2\right]-2.25=2.5-2.25=0.25[(1)2(0.5)+(2)2(0.5)]−(1.5)2=[0.5+2]−2.25=2.5−2.25=0.25

Variance (Method 3):

Var(X)=E(X2)−E(X)2Var\left(X\right)=E\left(X^2\right)-E\left(X\right)^2Var(X)=E(X2)−E(X)2

E(X)=(1)(0.5)+(2)(0.5)=1.5E\left(X\right)=\left(1\right)\left(0.5\right)+\left(2\right)\left(0.5\right)=1.5E(X)=(1)(0.5)+(2)(0.5)=1.5 (solved above)

E(X2)=[(1)2(0.5)+(2)2(0.5)]=0.5+2=2.5E\left(X^2\right)=\left[\left(1\right)^2\left(0.5\right)+\left(2\right)^2\left(0.5\right)\right]=0.5+2=2.5E(X2)=[(1)2(0.5)+(2)2(0.5)]=0.5+2=2.5 (solved above)

So, Var(X)=E(W2)−E(W)2=2.5−(1.5)2=2.5+2.25=0.25Var\left(X\right)=E\left(W^2\right)-E\left(W\right)^2=2.5-\left(1.5\right)^2=2.5+2.25=0.25Var(X)=E(W2)−E(W)2=2.5−(1.5)2=2.5+2.25=0.25

Method 3 is really just the same as Method 2 but broken down into more steps.

Standard deviation:

SD(X)=σ=Var(X)SD(X)=\sigma=\sqrt{Var(X)}SD(X)=σ=Var(X)​

SD(X)=0.25=0.5SD\left(X\right)=\sqrt{0.25}=0.5SD(X)=0.25​=0.5

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Example with 2 random variables X and Y:

xyP(X∩Y)260.2480.8\begin{array}{|c|c|c|}

\hline
x&y&P(X\cap Y)\\
\hline
2&6&0.2\\
\hline
4&8&0.8\\
\hline
\end{array}x24​y68​P(X∩Y)0.20.8​​

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Example: Random Variables & Joint Distribution
xyP(X∩Y)260.2480.8\begin{array}{|c|c|c|}

\hline
x&y&P(X\cap Y)\\
\hline
2&6&0.2\\
\hline
4&8&0.8\\
\hline
\end{array}x24​y68​P(X∩Y)0.20.8​​

E(XY)=E\left(XY\right)=E(XY)=

Let W=XYW=XYW=XY

E(XY)→E(W)=(2×6)(0.2)+(4×8)(0.8)=2.4+25.6=28E\left(XY\right)\rightarrow E\left(W\right)=\left(2\times6\right)\left(0.2\right)+\left(4\times8\right)\left(0.8\right)=2.4+25.6=28E(XY)→E(W)=(2×6)(0.2)+(4×8)(0.8)=2.4+25.6=28

 Var(XY)=Var\left(XY\right)=Var(XY)=

Let W=XYW=XYW=XY
Var(W)=E(W2)−E(W)2Var\left(W\right)=E\left(W^2\right)-E\left(W\right)^2Var(W)=E(W2)−E(W)2

E(W)=28E(W)=28E(W)=28 

E(W2)=(2×6)2(0.2)+(4×8)2(0.8)=28.8+819.2=848E\left(W^2\right)=\left(2\times6\right)^2\left(0.2\right)+\left(4\times8\right)^2\left(0.8\right)=28.8+819.2=848E(W2)=(2×6)2(0.2)+(4×8)2(0.8)=28.8+819.2=848

∴Var(XY)→ Var(W)=E(W2)−E(W)2=848−(28)2=64\therefore Var\left(XY\right)\rightarrow\ Var\left(W\right)=E\left(W^2\right)-E\left(W\right)^2=848-\left(28\right)^2=64∴Var(XY)→ Var(W)=E(W2)−E(W)2=848−(28)2=64

SD(XY)=SD\left(XY\right)=SD(XY)=

=Var(XY)=64=8=\sqrt{Var\left(XY\right)}=\sqrt{64}=8=Var(XY)​=64​=8

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Practice: Random Variables & Joint Distribution

Let X and Y  be two discrete random variables with the following joint probability distribution function: 

xyP(X∩Y)101/3212/3\begin{array}{|c|c|c|}

\hline
x&y&P(X\cap Y)\\
\hline
1&0&1/3\\
\hline
2&1&2/3\\
\hline
\end{array}x12​y01​P(X∩Y)1/32/3​​


(a) Find E(X)E(X)E(X) 

(b) Find E(Y)E(Y)E(Y) 

(c) Find E(XY)E(XY)E(XY) 

(d) Find Var(XY)Var(XY)Var(XY)

I have answered this question

Example: Random Variables & Joint Distribution
Let X and Y  be two discrete random variables with the following joint probability distribution function: 

xyP(X∩Y)101/3212/3\begin{array}{|c|c|c|}

\hline
x&y&P(X\cap Y)\\
\hline
1&0&1/3\\
\hline
2&1&2/3\\
\hline
\end{array}x12​y01​P(X∩Y)1/32/3​​



Find SD(2X−Y)SD(2X-Y)SD(2X−Y)

Let 𝑊=2𝑋−𝑌𝑊 = 2𝑋 − 𝑌W=2X−Y. Then we get the following pdf:
wP(W=w)\begin{array}{c|c}
w&P(W=w)\\
\hline 
\\
\\

\end{array}w​P(W=w)​

wP(W=w)2(1)−0=21/32(2)−1=32/3\begin{array}{c|c}
w&P(W=w)\\
\hline
2(1)-0=2&1/3\\
2(2)-1=3&2/3
\end{array}w2(1)−0=22(2)−1=3​P(W=w)1/32/3​​


E(W)=(2)(13)+(3)(23)=83E(W)=\left(2\right)\left(\frac{1}{3}\right)+\left(3\right)\left(\frac{2}{3}\right)=\frac{8}{3}E(W)=(2)(31​)+(3)(32​)=38​

E(W2)=(2)2(13)+(3)2(23)=223E\left(W^2\right)=\left(2\right)^2\left(\frac{1}{3}\right)+\left(3\right)^2\left(\frac{2}{3}\right)=\frac{22}{3}E(W2)=(2)2(31​)+(3)2(32​)=322​

So, Var(W)=E(W2)−E(W)2=223−(83)2=223−649=29Var\left(W\right)=E\left(W^2\right)-E\left(W\right)^2=\frac{22}{3}-\left(\frac{8}{3}\right)^2=\frac{22}{3}-\frac{64}{9}=\frac{2}{9}Var(W)=E(W2)−E(W)2=322​−(38​)2=322​−964​=92​

Therefore, SD(2X−Y)→SD(W)=29=23SD\left(2X-Y\right)\rightarrow SD\left(W\right)=\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}SD(2X−Y)→SD(W)=92​​=32​​

Practice: Random Variables & Joint Distribution
Let X and Y  be two discrete random variables with the following joint probability distribution function: 

xyP(X∩Y)240.6370.4\begin{array}{|c|c|c|}

\hline
x&y&P(X\cap Y)\\
\hline
2&4&0.6\\
\hline
3&7&0.4\\
\hline
\end{array}x23​y47​P(X∩Y)0.60.4​​

Assume independence.

(a) Find E(X+Y)E(X+Y)E(X+Y) 

(b) Find SD(4X−Y−1)SD(4X-Y-1)SD(4X−Y−1)

(a) Find E(X+Y). Provide answer with one decimal place (e.g. 5.4)

(b) Find SD(4X-Y-1). Provide answer with at least one decimal place (e.g. 2.6)

I don't know