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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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 Spring '09
 wen
 Polar Coordinates

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