(7) By using the mgf, show that the sum of independent chi-square random variables is again a chi-square random variable. That is, if U ~ x and V ~ xm and U,V are independent then show that U + V ~ xátm) The following proofs are proposed. (a) We know that E(e") = ent12, tE R. Next, if U, V are independent and U - and V ~ Xím) then the mgf of U + V is E(e«U+V)) = E(eU) E(e") = en²12 mi²12 = eln+m)r²12_ Therefore, U + V ~ Xntm): (b) The stated result is, in fact, false even though the mgfs of U and V exist (c) The stated result is, in fact, false since the mgf of U or V does not exist. (d) We know that E(e") = e tE R. Next, if U, V are independent and U ~ X and V ~ xm then the mgf of U + V is E(eU+V)) = E(e'U) E(eV) = e¯nt³12 e-mi*12 = e-(n+m)x*1n Therefore, U + V ~ Xntm): (e) We know that E") - (z). E(eU) = t < 0.5, V1 – 21 which is the mgf of X Next, if U, V are independent and U ~ X and V ~ xm then the mgf of U + V is n/2 m/2 (n+m)/2 E(e«U+V)) = E(eU) E(e") = 1- 2t 1– 2 Therefore, U + V ~ Xxtm): The correct answer is (a) (b) (c) (d) (e) N/A

(7) By using the mgf, show that the sum of independent chi-square random variables is again a chi-square random variable. That is, if U ~ x and V ~ xm and U,V are independent then show that U + V ~ xátm) The following proofs are proposed. (a) We know that E(e") = ent12, tE R. Next, if U, V are independent and U - and V ~ Xím) then the mgf of U + V is E(e«U+V)) = E(eU) E(e") = en²12 mi²12 = eln+m)r²12_ Therefore, U + V ~ Xntm): (b) The stated result is, in fact, false even though the mgfs of U and V exist (c) The stated result is, in fact, false since the mgf of U or V does not exist. (d) We know that E(e") = e tE R. Next, if U, V are independent and U ~ X and V ~ xm then the mgf of U + V is E(eU+V)) = E(e'U) E(eV) = e¯nt³12 e-mi12 = e-(n+m)x1n Therefore, U + V ~ Xntm): (e) We know that E") - (z). E(eU) = t < 0.5, V1 – 21 which is the mgf of X Next, if U, V are independent and U ~ X and V ~ xm then the mgf of U + V is n/2 m/2 (n+m)/2 E(e«U+V)) = E(eU) E(e") = 1- 2t 1– 2 Therefore, U + V ~ Xxtm): The correct answer is (a) (b) (c) (d) (e) N/A

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

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U, V
2.
are independent then show that U + V ~ Xntm):
(7) By using the mgf, show that the sum of independent chi-square random variables is again a chi-square random variable. That is, if U ~
and V
X(m)
and
The following proofs are proposed.
(a) We know that
E(eU) = enr*2_
tE R.
Next, if U, V are independent and U ~ X and V ~ xm then the mgf of U +V is
E(eU+V)) = E(e'U) E(e") = enr*12 emi²12 = eln+m)r*12_
Therefore, U + V ~ xntm).
(b) The stated result is, in fact, false even though the mgfs of U and V exist.
(c) The stated result is, in fact, false since the mgf of U or V does not exist.
(d) We know that
E(eU) = e¬n³12,
tE R.
Next, if U, V are independent and U~ X and V ~ Xím then the mgf of U + V is
E(e(U+V)) = E(eU) E(eV) = e¯nt²12 e-mi²12 = e-(n+m)r*12_
Therefore, U + V ~
.2
(n+m)*
(e) We know that
FI") = ().
n
1
t < 0.5,
2t
which is the mgf of Xín) : Next, if U, V
are independent and U ~ Xím and V ~
then the mgf of U+ V is
(프구)-.(구)..(무)
n/2
1
m/2
(n+m)/2
E(eU+V)) = E(eU) E(e") =
- 2t
1 – 2t
1 – 2t
.2
Therefore, U + V ~
Xīn+m)'
The correct answer is
(a)
(b)
(c)
(d)
N/A
(Select One)

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