Section II 11. Assume X is a random variable with moment generating function M(t) (0.3e" +0.7), -00
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![**Section II**
11. Assume \( X \) is a random variable with moment generating function
\[
M(t) = (0.3e^{4t} + 0.7)^3, \quad -\infty < t < \infty.
\]
Note that no special type of random variable we have studied thus far in MAT 353 has this moment generating function.
(a) Compute \( E(X) \).
(b) Compute \( \text{Var}(X) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F653b506a-bf1f-413c-8a44-59b8a78c6c88%2F15d68c5e-278b-413c-a0e1-6158548ddaa3%2Fmvf3wt_processed.jpeg&w=3840&q=75)
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- Part III: Sections 3.1- 3.3 13. Suppose that X is a continuous random variable with pdf given as 43 if 1< <4 255 f(1) = %3D otherwise. (a) For 12. Find the distribution of the random variables that have each of the following moment generating functions: a) m(t) = e2(e*–1) b) m(t) = (¿) e* + (¿). e2t + e3tif x be a random variable with moment generating function m,(t) = (0.6+ 0.4e*)10 then E(x)=7. A discrete random variable X taking non-negative integer values has probability gen- erating function: 1 rx(t) = %3D (2 – t)2 (b) Calculate E(X).Example 3.6. Let X and Y be jointly continuous random variables with joint PDF is given by: 3 fx.x(x, y) =(1+ a²y)!(0,1)(x)/(0.2)() Note that fx(x) = (1+ x²) I0,1) (2) fy(y) : (1+ y) I(0,2)(y) Solve for fx|y(x | y). Remark. It follows immediately from the definition of conditional probability density function that fx,y(r,y) = fxjy(x | y) · fy (y), fx.x (x, y) = fy|x(y | x) · fx(x), -003. find the distributions of the random variables that have each of the following moment generating functions: et 2-et 1 a) m(t) = = b) m(t) = e²(et-1) c) Using results in (a) and (b), Find the E(Y) and V (Y)? Suppose that Y is a random variable with moment-generating function m(t) If(4) Let X be a random variable with p.d.f. 2e-2* f(x) = 0Question 4 Let X be a random variable with AX) 0, EX) 0.6. if x 1 Answer (1) and (2): (1) If X, X2, X3, X4 are iid f, then (i) PX4) S0.8) = (.. ..). (ii) PX1) < 0) = (...). (2) Let X, X2,.., X36 be iid f You are given: PZS 1.96) = 0.975; P(ZS 2) = 0.977; P(Z< 1.55) 0.939, PZS 1.81) = 0.965. Use the CLT to approximate PXS 0.2). %3D %3D %3DProve the followingSEE MORE QUESTIONSRecommended textbooks for youA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON
A First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON