(10) For the above problem verify that the theorem of total variance (TTV) holds. The following verifications are proposed. (a) Since Var(Var(X|Y)) = o = Var(X), the TTV holds. (b) Since p*ožVar(Y) of +ož(1 – p²) = o = Var(X). Var(E(X|Y)) + E(Var(X|Y)) = the TTV holds. (c) Since Var(E(X|Y)) = Var(X) = oi, the TTV holds. (d) Since Var(E(X\Y)) = 0 and E(Var(X|Y)) = Var(X), the TTV holds. (e) None of the above The correct verification is (a) (b) (c) (d) (e) N/A (Select One)
(10) For the above problem verify that the theorem of total variance (TTV) holds. The following verifications are proposed. (a) Since Var(Var(X|Y)) = o = Var(X), the TTV holds. (b) Since p*ožVar(Y) of +ož(1 – p²) = o = Var(X). Var(E(X|Y)) + E(Var(X|Y)) = the TTV holds. (c) Since Var(E(X|Y)) = Var(X) = oi, the TTV holds. (d) Since Var(E(X\Y)) = 0 and E(Var(X|Y)) = Var(X), the TTV holds. (e) None of the above The correct verification is (a) (b) (c) (d) (e) N/A (Select One)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![(10) For the above problem verify that the theorem of total variance (TTV) holds.
The following verifications are proposed.
(a) Since Var(Var(X|Y)) = o = Var(X), the TTV holds.
(b) Since
poVar(Y)
Var(E(X|Y)) + E(Var(X|Y)) =
+ož(1– p²) = o = Var(X).
the TTV holds.
(c) Since Var(E(X|Y)) = Var(X) = o3, the TTV holds.
(d) Since Var(E(x|Y)) = 0 and E(Var(X|Y)) = Var(X), the TTV holds.
(e) None of the above
The correct verification is
(a)
(b)
(c)
(e)
N/A
(Select One)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F421db798-cca4-4efb-a6a9-3775d86708c7%2F0a48e7af-a365-43ba-aff3-4382d83782ea%2Fop74dgp_processed.png&w=3840&q=75)
Transcribed Image Text:(10) For the above problem verify that the theorem of total variance (TTV) holds.
The following verifications are proposed.
(a) Since Var(Var(X|Y)) = o = Var(X), the TTV holds.
(b) Since
poVar(Y)
Var(E(X|Y)) + E(Var(X|Y)) =
+ož(1– p²) = o = Var(X).
the TTV holds.
(c) Since Var(E(X|Y)) = Var(X) = o3, the TTV holds.
(d) Since Var(E(x|Y)) = 0 and E(Var(X|Y)) = Var(X), the TTV holds.
(e) None of the above
The correct verification is
(a)
(b)
(c)
(e)
N/A
(Select One)
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