# 4 example the expected value of the binomial

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Example. The expected value of the binomial distribution b ( x ; n, p ) is E ( x ) = X x xb ( x ; n, p ) = n X x =0 x n ! ( n x )! x ! p x (1 p ) n x . We factorise np from the expression under the summation and we begin the summation at x = 1. On defining y = x 1, which means setting x = y + 1 in the expression above, we get E ( x ) = n X x =0 xb ( x ; n, p ) = np n 1 X y =0 ( n 1)! ([ n 1] y )! y ! p y (1 p ) [ n 1] y = np n 1 X y =0 b ( y ; n 1 , p ) = np, where the final equality follows from the fact that we are summing the values of the binomial distribution b ( y ; n 1 , p ) over its entire domain to obtain a total of unity. 5
Example. Let x f ( x ) = e x ; 0 x < . Then E ( x ) = Z 0 xe x dx. This must be evaluated by integrating by parts: Z u dv dx dx = uv Z v du dx dx. With u = x and dv/dx = e x , this formula gives Z 0 xe x dx = [ xe x ] 0 + Z 0 e x dx = [ xe x ] 0 [ e x ] 0 = 0 + 1 = 1 . Observe that, since this integral is unity and since xe x > 0 over the domain of x , it follows that f ( x ) = xe x is also a valid p.d.f. 6

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Expectation of a function of random variable. Let y = y ( x ) be a function of x f ( x ). The value of E ( y ) can be found without first determining the p.d.f g ( y ) of y . Quite simply, there is E ( y ) = Z x y ( x ) f ( x ) dx. If y = y ( x ) is a monotonic transformation of x , then it follows that E ( y ) = Z y yg ( y ) dy = Z y yf { x ( y ) } Ø Ø Ø Ø dx dy Ø Ø Ø Ø dy = Z x y ( x ) f ( x ) dx, which establishes a special case of the result. However, the result is not confined to monotonic transformations. It is equally valid for functions that are piecewise monotonic, i.e. ones that
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• Spring '12
• D.S.G.Pollock
• Normal Distribution, Probability theory, probability density function, dx, Ø

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