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Metallideth
March 9th 2011, 06:55 PM
I'm completely lost on this one.

If X is N(μ,σ^2), show that Y = aX + b is N(aμ + b, (a^2)(σ^2)), a is not equal to 0. Hint: Find the distribution function P(Y less than y) of Y, and in the resulting integral, let w = ax + b, or, equivalently, x = (w-b)/a

If I do the substitution, the integral is taken from negative infinity to y. That's all fine and good, but then the expression above the "e" term becomes awful, and I see no way to simplify it.

NoMoreExams
March 9th 2011, 07:20 PM
F_{X}(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-(z - \mu)^2 / (2 \sigma^2)} dz

So now when you say Y = aX + b you have

Pr(Y <= y) = Pr(aX + b <= y) = Pr(X <= (y-b)/a) so plug it in

=> \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-((y-b)/a - \mu)^2 / (2 \sigma^2)} dy

So the trick is with working with the "stuff" in the exponent namely you can rewrite

((y-b)/a - \mu)^2/(2 \sigma^2) = (y - b - a \mu)^2 / (2a^2 \sigma^2)

So now compare what you had:

1) z - \mu now translates to y - (b + a \mu)

2) 2 \sigma^2 now translates to 2a^2 \sigma^2

Make sense?

1) tells you that the new mean is "b + a \mu"

2) tells you that the new variance is 2a \sigma^2