4700 proofs

If A and B are independent, so are A and B’

1. define A: (A and B) or (A and B’) → P(A) = P(A and B) + P(A and B’)

2. Isolate P(A and B’) = P(A) - P(A and B)

3. Factor: P(A and B’) = P(A)(1-P(B))

4. P(A and B’) = P(A)P(B’)

If 𝑎 and 𝑏 are constants, then 𝐸(𝑎𝑋 + 𝑏) = 𝑎𝐸(𝑋) + 𝑏. (discrete version)

1. E(aX+b) = sum(ax f(X)) + sum(b f(X)

2. a(sum(Xf(X))) + b(sum(f(X)))

3. aE(X) + b(1) = aE(X) + b

If 𝑎 and 𝑏 are constants, then 𝐸(𝑎𝑋 + 𝑏) = 𝑎𝐸(𝑋) + 𝑏. (continuous version)

1. E(aX+b) = int(aX f(X)dx) + int(b f(X)dx)

2. = a int(Xf(X)dx) + b int(f(X)dx)

3. aE(X) + b(1) = aE(X) + b

sigma² = mu’2 - mu²

1. E((X-mu))² = E(X² - 2Xmu - mu²)

2. = E(X²) - E(2Xmu) + E(mu²)

3. E(X²) - 2muE(X) + E(mu²)

4. E(X²) - 2(mu)(mu) + mu² = E(X²) - mu²

5. Set up LHS E(X²) - mu² = E(X²) - mu²

If X has the variance sigma², than var(aX+b) = a²sigma²

1. var(aX+b) = E((aX+b)²) - (E(aX + b))²

2. = E(a²X² + 2aXb + b²) - (aE(X) + b)²

3. = a²E(X²) +2abE(X) + b² - (a²(E(X))² + 2abE(X) + b²)

4. = a²(E(X²) - (E(X))²)

5. = a²(sigma²)

The moment-generating function of the Poisson distribution is given by Mx(t) = e^l(ambda * (e^t -1)

1. Mx(t) = sum(e^xt (p(x; lambda))

2. = e^-lambda sum( Mac series: (lambdae^t)^x/x!)

3. e^-lambda (e^z) = e^-lambda (e^lambda e^t)

4. e^(lambda)(e^t - 1)