We can also calculate the expected value and variance

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Unformatted text preview: We can also calculate the expected value and variance of a U (0 , 1) random variable using the formulas given in the previous section. For the expected value, we have E ( X ) = Z ∞-∞ xf X ( x )d x = Z b a x b- a d x = 1 2 b 2- 1 2 a 2 b- a = 1 2 ( a + b ) , which is just the midpoint of the interval ( a,b ). To obtain the variance, it is helpful to first calculate E ( X 2 ): E ( X 2 ) = Z ∞-∞ x 2 f X ( x )d x = Z b a x 2 b- a d x = 1 3 b 3- 1 3 a 3 b- a = 1 3 ( a 2 + ab + b 2 ) . Combining our expressions for E ( X 2 ) and E ( X ) yields Var( X ) = E ( X 2 )- E ( X ) 2 = 1 3 ( a 2 + ab + b 2 )- 1 2 ( a + b ) 2 = 1 12 ( b- a ) 2 . 4 The standard normal distribution The standard normal distribution is another very important example of a con- tinuous distribution. A random variable X is said to have the standard normal distribution, written X ∼ N (0 , 1), if it is continuous with pdf f X given by f X ( x ) = 1 √ 2 π e- x 2 / 2 . The graph of this function is the well-known “bell curve”. One can show that the area under the bell curve is equal to one, but this is a little tricky and we will not provide details. The numbers 0 and 1 in the expression N (0 , 1) refer to the expected value and variance of the standard normal distribution, which are zero and one respectively. We can show that the expected value is zero in the following way: E ( X ) = Z ∞-∞ x · 1 √ 2 π e- x 2 / 2 d x =- 1 √ 2 π Z ∞-∞ (- x ) · e- x 2 / 2 d x =- 1 √ 2 π e- x 2 / 2 ∞-∞ =- 1 √ 2 π (0- 0) = 0 . 4 To show that the variance of the standard normal distribution is one, we use a slightly more complicated argument involving integration by parts: Var( X ) = E ( X 2 ) = Z ∞-∞ x 2 · 1 √ 2 π e- x 2 / 2 d x = Z ∞-∞- x √ 2 π · - x e- x 2 / 2 d x =- x √ 2 π e- x 2 / 2 ∞-∞ + Z ∞-∞ 1 √ 2 π e- x 2 / 2 d x = (0- 0) + 1 = 1 . The standard normal distribution is referred to as “standard” because it has mean zero and variance one. More generally, we say that X is normally distributed with mean μ and variance σ 2 , written X ∼ N ( μ,σ 2 ), if it is continuous with pdf f X ( x ) = 1 √ 2 πσ 2 e- ( x- μ ) 2 / 2 σ 2 . One can show, using arguments similar to those used in the previous paragraph, that this formula for f X does indeed yield E ( X ) = μ and Var( X ) = σ 2 . A very useful property of the normal distribution is that a linear function of a normal random variable is still normal, but with different mean and variance. That is, if X ∼ N ( μ,σ 2 ), then a + bX ∼ N ( a + bμ,b 2 σ 2 ). This property of the normal distribution allows us to “standardize” a normal random variable by subtracting its mean and dividing by its standard deviation. The standardized random variable will then have the standard normal distribution....
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We can also calculate the expected value and variance of a...

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