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LEC15 Mean and Var of Cont rv

# LEC15 Mean and Var of Cont rv - V X b a =-2 12 4 Example...

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Expected Value of a Continuous Random Variable The expected value or mean of a continuous random variable with density f(x) is E X xf x dx ( ) ( ) = = -∞ μ ex: Find E(X) given f x e x x ( ) = - λ λ 0 . ( 29 E X x dx x e dx xe e dx e x x x x ( ) = + = - - - = - = - - = -∞ - - - - 0 0 0 0 1 1 1 0 0 0 λ λ λ λ λ λ λ λ using integration by parts: udv uv vdu = - where u x dv e dx du dx v e x x = = - - = = - 2 1 2 2

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The Expected Value of a Function of a Continuous Random Variable Let X be continuous random variable with density f(x). The expected value of the function h(X) is E h X h x f x dx ( ( )) ( ) ( ) = -∞ The Variance of a Continuous Random Variable Let X be continuous random variable with density f(x). The variance of X is [ ] V X E X E X E X x f x dx ( ) ( ) ( ) ( ) ( ) = - = - = - -∞ μ μ 2 2 2 2 2 2
Linear Functions of Continuous Random Variables If h(x) is a linear function of the form h(x)=a+bX then E(h(X)) = a + bE(X) V(h(x)) = b 2 V(X) Example: The Uniform Distribution on the Interval (a,b) Write f(x) Draw the density Give F(x) Find E(X)= (a+b)/2. 3

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Find E(X 2 ) = 1 3 2 2 + ( ) a b Find

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Unformatted text preview: V X b a ( ) =-2 12 4 Example: Mean & Variance of a function of a random variable The lifetime of a battery is exponentially distributed with mean 2; i.e, f x e x x ( ) . . . = ≥-5 5 E(X)=2 and V(X)=4 • The cost to operate the battery is a function of X: C=5+3x. Find the mean and variance of the cost. • The cost to operate the battery is D=3x 2 . Find the mean of the cost. 5 Percentiles The p th percentile of the r.v. X is the number such that a p percent of the distribution is less than that number The life of a component has the exponential distribution E(X)=1; i.e., f x e x x ( ) = ≥-Find the 50 th percentile (the median) = .69 Find 99 th percentile = 4.6 50 percentile: Find m such that e dx e e m m x m m m---= ⇔-= ⇔ = ⇔-= ⇔ = ∫ . . . ln. . 5 1 5 5 5 1 69 Note: for exponential mean ≠ median 6 7...
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