elemprob-page16 - A synonym for normal is Gaussian. The...

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Unformatted text preview: A synonym for normal is Gaussian. The first thing to do is show that this is a density. Let I = Then ∞ ∞ 2 e−x I2 = /2 −y 2 /2 e ∞ −x2 /2 e dx. 0 dx dy. −∞ 0 Changing to polar coordinates, ∞ π /2 re−r I2 = So I = π /2, hence Note 2 ∞ e−x /2 dx −∞ = √ 2 /2 dr = π/2. 0 0 2π as it should. 2 xe−x /2 dx = 0 by symmetry, so E Z = 0. For the variance of Z , we use integration by parts: 1 E Z2 = √ 2π 2 x2 e−x /2 1 dx = √ 2π 2 x · xe−x The integral is equal to 2 −xe−x /2 ∞ −∞ 2 e−x + /2 dx = √ /2 dx. 2π. Therefore Var Z = E Z 2 = 1. We say X is a N (µ, σ 2 ) if X = σZ + µ, where Z is a N (0, 1). We see that FX (x) = P(X ≤ x) = P(µ + σZ ≤ x) = P(Z ≤ (x − µ)/σ ) = FZ ((x − µ)/σ ) if σ > 0. (A similar calculation holds if σ < 0.) Then by the chain rule X has density fX (x) = FX (x) = FZ ((x − µ)/σ ) = 1 fZ ((x − µ)/σ ). σ This is equal to √ 2 2 1 e−(x−µ) /2σ . 2πσ E X = µ + E Z and Var X = σ 2 Var Z , so E X = µ, Var X = σ 2 . If X is N (µ, σ 2 ) and Y = aX + b, then Y = a(µ + σZ ) + b = (aµ + b) + (aσ )Z , or Y is N (aµ + b, a2 σ 2 ). In particular, if X is N (µ, σ 2 ) and Z = (X − µ)/σ , then Z is N (0, 1). The distribution function of a standard N (0, 1) is often denoted Φ(x), so that 1 Φ(x) = √ 2π x e−y 2 /2 dy. −∞ Tables of Φ(x) are often given only for x > 0. One can use the symmetry of the density function to see that Φ(−x) = 1 − Φ(x); this follows from −x Φ(−x) = P(Z ≤ −x) = −∞ ∞ = x 2 1 √ e−y /2 dy 2π 2 1 √ e−y /2 dy = P(Z ≥ x) = 1 − P(Z < x) = 1 − Φ(x). 2π 16 ...
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