15. This problem extends Problem 20.6. Let X, Y be random variables with finite mean. Show that 00 (P(X ≤ x ≤ Y) - P(X ≤ x ≤ X))dx = E Y — E X.
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mean. Show that
00
(P(X ≤ x ≤ Y) - P(X ≤ x ≤ X))dx = E Y — E X."
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Step by step
Solved in 2 steps with 4 images
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- Repeat Example 5 when microphone A receives the sound 4 seconds before microphone B.15. This problem extends Problem 20.6. Let X, Y be random variables with finite mean. Show that (P(X ≤ x ≤ Y) - P(Y < x ≤ X))dx = E Y — E X.3. Let the random variable X have the pmf f(x) = a) E(X) b) E(X²) c) E(3X²2X + 4) (x+1)² 9 for x = -1,0,1. Compute
- Suppose X is random variable whose p.d.f. is f2)=(2x-x²), 0, OSxs2 Suppose X is random variable whose p.d.f. is f(x)={4 elsewhere Find the mode , if it exists.4. Let X₁, X₂ be i.i.d. Exponential (A) random variables. Define { Z = I(X₁ + X₂ > 7) = Find E[Z|X₁] and E[Z] = E[E(Z|X₁)]. 1, if X₁ +X₂ > 7, 0, if X₁ + X₂ ≤ 7.2. Suppose one has n non-degenerate random variables X1, X2,, X, so that X1+ X2 + · · · + Xn = L for some constant L. (Recall that a random variable is non-degenerate if it is not a constant in disguise.) Show that there must be at least one pair of indices i j so that p(X;, X;) < 0.
- 6. Determine the constant c so that f(x)= cx, x-1, 2, 3,....10 satisfies the conditions of being a p. m. f. for a random variable X.4. (a) Consider a nonnegative, integer valued random variable Y , and a random sample X1,., Xn. + Xy) in terms of E(X1) Assume Y and all the X;'s are independent. Find E(X1+ and E(Y). ... (b) Suppose each person who logs onto Amazon.com on Christmas Eve is expected to spend $80. One hundred people are randomly chosen to see how much is spent by them. Each of them visits Amazon.com with probability 60%. How much money do we expect this group of people to spend at Amazon.com?4. If the random variable X has an exponential distrib show that for a > 0, b>0, 1 f(x) = -x/², for x ≥ 0, e 2 P(X> a+b|x> a) = P(X > b).
- 6. Let, for p = (0, 1), and xe R. X be a random variable defined as follows: P(X=-x) = P(X = x)=p. P(X=0)= 1-2p. Show that there is equality in Chebyshev's inequality for X. This means that Chebyshev's inequality, in spite of being rather crude, cannot be improved without additional assumptions.4. Suppose X1, X2,..., X10,000 are independent random variables with E(X;) = 8 and Var(X;) = 4 for each i. Let Y = (X1+ X2 + ... +X10,000)/10,000. a. Find E(Y). b. Find Var(Y). c. What does the Chebyshev inequality say about P(|Y – E(Y)| 2 0.1)?2.3.7. Suppose X1 and X2 are discrete random variables which have the joint pmf p(x1, r2) = (3x1+r2)/24, (21, 2) = (1, 1), (1, 2), (2, 1), (2, 2), zero elsewhere. Find the conditional mean E(X2 1), when r1 = 1.
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