MHB Find μ and σ^2 of Y when X~N(2,4)

  • Thread starter Thread starter dlee
  • Start date Start date
AI Thread Summary
If X follows a normal distribution X ~ N(2,4), then the transformed variable Y = -X - 1 is also normally distributed as Y ~ N(μ,σ^2). The expected value E(Y) is calculated as E(Y) = -E(X) - 1, resulting in μ = -3. The variance of Y remains the same as that of X, which is σ^2 = 4, since the variance is unaffected by adding or subtracting a constant. Therefore, the parameters for Y are μ = -3 and σ^2 = 4.
dlee
Messages
4
Reaction score
0
If X ~ N(2,4) then Y = -X-1 is also a normal random variable such that Y ~ N(μ,σ^2).
Find μ and σ^2.

I know that E(X) = 2 and Var(X) = 4.

E(Y) = -E(X) - 1
E(Y) = -2 - 1 = -3
So I found μ, but I'm not sure how to find the variance. Help?
 
Mathematics news on Phys.org
dlee said:
If X ~ N(2,4) then Y = -X-1 is also a normal random variable such that Y ~ N(μ,σ^2).
Find μ and σ^2.

I know that E(X) = 2 and Var(X) = 4.

E(Y) = -E(X) - 1
E(Y) = -2 - 1 = -3
So I found μ, but I'm not sure how to find the variance. Help?

Welcome to MHB, dlee! :)

There are a couple of basic properties for expectations and variances as you can see here.

In particular $\sigma^2(X + a)=\sigma^2(X)$ and $\sigma^2(aX)=a^2\sigma^2(X)$, where $a$ is some arbitrary constant.
 
AH the variance is 4! Thank you!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top