Chapter 5: JOINT PROBABILITY DISTRIBUTIONS Part 3:...
Transcript of Chapter 5: JOINT PROBABILITY DISTRIBUTIONS Part 3:...
Chapter 5: JOINT PROBABILITYDISTRIBUTIONS
Part 3:The Bivariate Normal
Section 5-3.2
Linear Functions of Random VariablesSection 5-4
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The bivariate normal is kind of nifty because...
• The marginal distributions of X and Y areboth univariate normal distributions.
• The conditional distribution of Y given X isa normal distribution.
• The conditional distribution of X given Y isa normal distribution.
• Linear combinations of X and Y (such asZ = 2X + 4Y ) follow a normal distribution.
• It’s normal almost any way you slice it.
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•Bivariate Normal ProbabilityDensity Function
The parameters: µX , µY , σX , σY , ρ
fXY (x, y) =1
2πσXσY√
(1− ρ2)×
exp
{−1
2(1− ρ2)
[(x− µX)2
σ2X
− 2ρ(x− µX)(y − µY )
σXσY+
(y − µY )2
σ2Y
]}
for −∞ < x <∞ and −∞ < y <∞, with
parameters σX > 0 , σY > 0 ,−∞ < µX <∞, −∞ < µY <∞,
and −1 < ρ < 1.
Where ρ is the correlation between X and Y .
The other parameters are the needed parame-ters for the marginal distributions of X and Y .
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•Bivariate Normal
When X and Y are independent, the con-tour plot of the joint distribution looks like con-centric circles (or ellipses, if they have differentvariances) with major/minor axes that are par-allel/perpendicular to the x-axis:
The center of each circle or ellipse is at (µX , µY ).
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•Bivariate Normal
When X and Y are dependent, the contourplot of the joint distribution looks like concen-tric diagonal ellipses, or concentric ellipses withmajor/minor axes that are NOT parallel/perp-endicular to the x-axis:
The center of each ellipse is at (µX , µY ).
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•Marginal distributions of X and Y inthe Bivariate Normal
Marginal distributions of X and Y are nor-mal:
X ∼ N(µX , σ2X) and Y ∼ N(µY , σ
2Y )
Know how to take the parameters from thebivariate normal and calculate probabilitiesin a univariate X or Y problem.
•Conditional distribution of Y |x in theBivariate Normal
The conditional distribution of Y |x is alsonormal:
Y |x ∼ N(µY |x, σ2Y |x)
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Y |x ∼ N(µY |x, σ2Y |x)
where the “mean of Y |x” or µY |x dependson the given x-value as
µY |x = µY + ρσYσX
(x− µX)
and “variance of Y |x” or σ2Y |x depends on
the correlation as
σ2Y |x = σ2
Y (1− ρ2).
Know how to take the parameters from thebivariate normal and get a conditional distri-bution for a given x-value, and then calculateprobabilities for the conditional distributionof Y |x (which is a univariate distribution).
Remember that probabilities in the normalcase will be found using the z-table.
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Notice what happens to the joint distribution(and conditional) as ρ gets closer to +1:
ρ = 0.45
ρ = 0.75
ρ = 0.95
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As a last note on the bivariate normal...
Though ρ = 0 does not mean X and Y are in-dependent in all cases, for the bivariate normal,this does hold.
For the Bivariate Normal,Zero Correlation Implies Independence
If X and Y have a bivariate normal distribution(so, we know the shape of the joint distribution),then with ρ = 0, we have X and Y as indepen-dent.
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• Example: From book problem 5-54.
Assume X and Y have a bivariate normaldistribution with..µX = 120, σX = 5µY = 100, σY = 2ρ = 0.6
Determine:(i) Marginal probability distribution of X .
(ii) Conditional probability distribution of Ygiven that X = 125.
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Linear Functions of Random VariablesSection 5-4
• Linear Combination
Given random variables X1, X2, . . . , Xp andconstants c1, c2, . . . , cp,
Y = c1X1 + c2X2 + · · · + cpXp
is a linear combination of X1, X2, . . . , Xp.
•Mean of a Linear Function
If Y = c1X1 + c2X2 + · · · + cpXp,
E(Y ) = c1E(X1)+c2E(X2)+· · ·+cpE(Xp)
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•Variance of a Linear Function
If X1, X2, . . . , Xp are random variables, andY = c1X1 + c2X2 + · · ·+ cpXp, then
in general
V (Y ) = c21V (X1)+c2
2V (X2) + · · · + c2pV (Xp)
+ 2∑i<j
∑cicjcov(Xi, Xj)
In this class, all our linear combinations ofrandom variables will be done with indepen-dent random variables.
If X1, X2, . . . , Xp are independent,
V (Y ) = c21V (X1)+c2
2V (X2)+· · ·+c2pV (Xp)
The most common mistake for finding thevariance of a linear combination is to forgetto square the coefficients.
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• Example: Semiconductor product (exam-ple 5-31)
A semiconductor product consists of threelayers. The variance of the thickness of thefirst, second, third layers are 25, 40, and 30nanometers2.
What is the variance of the thickness of thefinal product?
ANS:
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•Mean and Variance of an Average
If X̄ =(X1+X2+···+Xp)
p
with E(Xi) = µ for i = 1, 2, . . . , p then
E(X̄) = µ.
The expected value of the average of p ran-dom variables, all with the same mean µ,is just µ again.
If X1, X2, . . . , Xp are also independent with
V (Xi) = σ2 for i = 1, 2, . . . , p then
V (X̄) =σ2
p
The variance of the average of p randomvariables is smaller than the variance of asingle random variable.
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•Reproductive Property of theNormal Distribution
If X1, X2, . . . , Xp are independent, normalrandom variables with E(Xi) = µi andV (Xi) = σ2
i for i = 1, 2, . . . , p,
Y = c1X1 + c2X2 + · · · + cpXp
is a normal random variable with
µY = E(Y ) = c1µ1 + c2µ2 + · · · + cpµp
and
σ2Y = V (Y ) = c2
1σ21 + c2
2σ22 + · · · + c2
pσ2p
i.e. Y ∼ N(µY , σ2Y ) as described above.
A linear combination of normal r.v.’s is also normal.
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• Example: Weights of people
Assume that the weights of individuals areindependent and normally distributed witha mean of 160 pounds and a standard devi-ation of 30 pounds. Suppose that 25 peoplesqueeze into an elevator that is designed tohold 4300 pounds.
a) What is the probability that the load ex-ceeds the design limit?
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