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18.44 Lecture Notes - Lecture 26: Random Variable, Royal Institute Of Technology, Weighted Arithmetic Mean

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amethystmuskrat145

22 May 2016

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Scott Sheffield

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So m(t) is a weighted average of countably many exponential functions. The moment generating function of x is de ned by m(t) = m (t) := e[e ] When x is discrete, can write m(t) = e p (x). When x is continuous, can write m(t) = e f(x) dx. If b > 0 and t > 0 then e[e ] e[e ] p{x b}e. If x takes both positive and negative values with positive probability then m(t) grows at least exponentially fast in |t| as |t| . So m(t) is a weighted average of a continuum of exponential functions. Let x be a random variable and m(t) = e[e ] Then m (t) = / t e[e ] = e[ / t (e )] = e[xe ] Also m (t) = / t m (t) = / t e[xe ] = e[x e ].

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Related Questions

5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

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