12.2 Covariance If we are interested at how two random variables vary together, we need to look at the covariance. Definition 12.1 Let X and Y be two random variables with expectations EX respectively. Then their covariance is Cov(X,Y)=E( X − +x)(Y – Uy). = μχ and ΕΥ = μy In the least surprising result of this whole module, we also have a computational formula to go along with this definitional formula. Theorem 12.2 Let X and Y be two random variables with expectations μx and μy respectively. Then their covariance can also be calculated as Cov(X,Y) =EXY – AxMy. Proof. Exactly as we've done many times before, we have Cov(X,Y)= E(X − &x)(Y – My) == = (ΧΥ - Χ μγ - μχΥ + μχμχ) =EXY-μy EX − μx EY + μ x My = EXY - μx My - Mx My + MX MY =EXY - UxUY, and we're done. B1. (a) Let X ~ Bin(n, p). By using the formula EX = Σxpx(x), x Show that EX = np. You may use the binomial expansion, Hint: First show that m Σπ agm-k (a+b)m = Σ k=0 (m) (+-) - (2)² }) (b) Let X and Y be random variables, and let a be a constant. Starting from the definition of covariance, show that Cov(aX,Y) = a Cov(X, Y). (c) Let X and Y be Bernoulli ( ✓ ✓) random variables. Write down a table for the joint PMF of X and Y for which X and Y are uncorrelated.
12.2 Covariance If we are interested at how two random variables vary together, we need to look at the covariance. Definition 12.1 Let X and Y be two random variables with expectations EX respectively. Then their covariance is Cov(X,Y)=E( X − +x)(Y – Uy). = μχ and ΕΥ = μy In the least surprising result of this whole module, we also have a computational formula to go along with this definitional formula. Theorem 12.2 Let X and Y be two random variables with expectations μx and μy respectively. Then their covariance can also be calculated as Cov(X,Y) =EXY – AxMy. Proof. Exactly as we've done many times before, we have Cov(X,Y)= E(X − &x)(Y – My) == = (ΧΥ - Χ μγ - μχΥ + μχμχ) =EXY-μy EX − μx EY + μ x My = EXY - μx My - Mx My + MX MY =EXY - UxUY, and we're done. B1. (a) Let X ~ Bin(n, p). By using the formula EX = Σxpx(x), x Show that EX = np. You may use the binomial expansion, Hint: First show that m Σπ agm-k (a+b)m = Σ k=0 (m) (+-) - (2)² }) (b) Let X and Y be random variables, and let a be a constant. Starting from the definition of covariance, show that Cov(aX,Y) = a Cov(X, Y). (c) Let X and Y be Bernoulli ( ✓ ✓) random variables. Write down a table for the joint PMF of X and Y for which X and Y are uncorrelated.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.2: Properties Of Division
Problem 50E
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Just do the part b, thank you so much
Transcribed Image Text:12.2 Covariance
If we are interested at how two random variables vary together, we need to look at the covariance.
Definition 12.1 Let X and Y be two random variables with expectations EX
respectively. Then their covariance is
Cov(X,Y)=E( X − +x)(Y – Uy).
=
μχ and ΕΥ = μy
In the least surprising result of this whole module, we also have a computational formula to go along
with this definitional formula.
Theorem 12.2 Let X and Y be two random variables with expectations μx and μy respectively.
Then their covariance can also be calculated as
Cov(X,Y) =EXY – AxMy.
Proof. Exactly as we've done many times before, we have
Cov(X,Y)= E(X − &x)(Y – My)
==
= (ΧΥ - Χ μγ - μχΥ + μχμχ)
=EXY-μy EX − μx EY + μ x My
= EXY - μx My - Mx My + MX MY
=EXY - UxUY,
and we're done.
Transcribed Image Text:B1.
(a) Let X
~
Bin(n, p). By using the formula
EX = Σxpx(x),
x
Show that EX
= np.
You may use the binomial expansion,
Hint: First show that
m
Σπ agm-k
(a+b)m = Σ
k=0
(m)
(+-) - (2)²
})
(b) Let X and Y be random variables, and let a be a constant. Starting from the definition of
covariance, show that Cov(aX,Y) = a Cov(X, Y).
(c) Let X and Y be Bernoulli ( ✓ ✓) random variables. Write down a table for the joint PMF of X and Y
for which X and Y are uncorrelated.
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