Statistics, Exponential Distribution

Question

Given an exponential distribution Y with mean 10, find the mean and variance of 𝐶=100+40𝑌+3𝑌² .

Answer 1

For an exponential distribution with mean 10, f(y) =1/10 e^-y/10 , and E(Yⁿ)= n! * 10ⁿ (this may easily be shown via integration by parts.

C= 100 + 40Y + 3Y², so

E(C) = 100 +40E(Y) + 3 E(Y²) = 100 + 40 * 1! * 10¹ + 3 * 2! * 10² =

100 + 40 * 10 + 6 * 100 =

1100

To calculate the variance, we need to determine E(C²).

(100 +40Y +3Y²)²=

100² + (2*100*40)Y + (40² + 2 * 100* 3)Y² + 2*40*3Y³ +9Y⁴

Then, E(C²) =

10⁴ + (2*4*10³)(10) + (4² + 2*3)10²(2*10²)+ 2*4*3*10* 3!*10³ + 9 * 4!* 10⁴ =

10⁴ (1 + 8 + 44 + 144 + 216) = 413 * 10^4

Then, var(C) =E(C²⁾ - E(C)² = 413 * 10⁴ - 11² * 10⁴= 292 * 10⁴