Prove that E (aX+b) =aE(X) +b?

Answer 1

Recall: If X is a discrete random variable with probability function P(X = x_i) = p(x_i), then the expected value of the random variable X is given by:

μ_x = E(X) = ∑ x_i• p(x_i)

i

So, E(aX + b) = ∑(ax_i + b) • p(x_i)

i

= a ∑x_i•p(x_i) + b ∑ p(x_i)

i i

= a•E(X) + b (Since ∑ p(x_i) represents the sum of all the probabilities which is 1)

i

Hope this helps

Mr. K

Answer 2

The expectation value can be represented as the integral of function over the measure...

Here we represent the measure μ(x) = a weighting function

E(aX+b) = ∫[lb=-∞][ub=∞] (aX+b) dμ / ∫[lb=-∞][ub=∞] dμ

Now the denominator integral ∫[lb=-∞][ub=∞] dμ is the normalization factor. Looking at the numerator integral we can use the associativity of integrals to break this up...

∫[lb=-∞][ub=∞] (aX+b) dμ = ∫[lb=-∞][ub=∞] (aX) dμ + ∫[lb=-∞][ub=∞] b dμ

Since a and b are constants...

∫[lb=-∞][ub=∞] (aX) dμ + ∫[lb=-∞][ub=∞] b dμ = a * ∫[lb=-∞][ub=∞] Xdμ + b * ∫[lb=-∞][ub=∞] dμ

Placing back into the definition...

∫[lb=-∞][ub=∞] (aX+b) dμ = a * ∫[lb=-∞][ub=∞] Xdμ + b * ∫[lb=-∞][ub=∞] dμ

Here we define ∫[lb=-∞][ub=∞] Xdμ = normalization factor (see above) * E(X) [ the expectation of value X ]

E(aX+b) = a * ∫[lb=-∞][ub=∞] Xdμ + b * ∫[lb=-∞][ub=∞] dμ / ∫[lb=-∞][ub=∞] dμ

E(aX+b) = a * ∫[lb=-∞][ub=∞] Xdμ / ∫[lb=-∞][ub=∞] dμ + b * ∫[lb=-∞][ub=∞] dμ / ∫[lb=-∞][ub=∞] dμ

E(aX+b) = a * E(X) + b