For p xed with 0 p 1 and any constant c lim p n snp

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Unformatted text preview: the density is symmetric about the point µ. To check that σ 2 is indeed the variance of the N (µ, σ 2 ) density, first note that if X is a standard normal random variable, then LOTUS and integration by parts yields ∞ u2 u2 √ exp − du 2 2π −∞ ∞ 1 u2 √ u · u exp − = du 2 2π −∞ ∞ 1 u u2 ∞ u2 √ exp − √ exp − + =− 2 2 2π 2π −∞ −∞ E [X 2 ] = du = 0 + 1 = 1. Since X has mean zero, the variance, Var(X ), for a standard normal random variable is one. Finally, a N (µ, σ 2 ) random variable Y can be written as σX + µ, where X is a standard normal random variable, so by the scaling formula for variances, Var(Y ) = σ 2 Var(X ) = σ 2 , as claimed. Example 3.6.4 Let X be a N (10, 16) random variable (i.e. a Gaussian random variable with mean 10 and variance 16). Find the numerical values of the following probabilities: P {X ≥ 15}, P {X ≤ 5}, P {X 2 ≥ 400}, and P {X = 2}. Solution: The idea is to use the fact that either the Φ or Q function. X −10 4 is a standard normal random...
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