Explore BrainMass

# Expectation of normal distribution

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

I know this problem is very easy, finding E(x) from the distribution function. I tried to do it by integration by parts, one way I took x to be my first function and the whole exp term to be my 2nd function, but it didn't work, then I split the exponential term to 2 terms combined one with x as my first function and took the second as the 2nd function but it didn't work either. Can someone help me?

See attached file for full problem description.

https://brainmass.com/statistics/normal-distribution/expectation-normal-distribution-78092

#### Solution Preview

.f(x) = __1__ exp-( x - μ )^2
√2πσ^2 2σ^2

α
E(x) = ∫ x __1__ exp-(x - μ)^2 dx
-α √2πσ^2 2σ^2

let ( x - μ ) = t ; dx = dt ; dx = √2σ dt
√2σ √2σ

x = √2σ t + μ
Substituting in the above integral we get
α
∫ (√2σ t + μ) { __1__ exp(- t^2 ) }√2σdt
-α √2πσ^2

α
∫ (√2σ t + μ) { __1__ ...

#### Solution Summary

The expert examines expectations of normal distributions.

\$2.19