Expected Value Variance Please Help!!

Question

Let 𝑋 be a random variable with density

𝑓(𝑥)={𝐶(4𝑥−𝑥^2) , 0≤𝑥≤3

0 , otherwise

where 𝐶∈𝐑 is some constant.

C = 1/9

E[X] = 1.75

Find E[x^2] (the second order moment of 𝑋) = ??

Find var[𝑋], the variance of the random variable 𝑋 = ??

Answer 1

To find the second order moment of 𝑋, we need to compute 𝐸[𝑋^2]:

𝐸[𝑋^2] = ∫(𝑥^2𝑓(𝑥))𝑑𝑥

Since 𝑓(𝑥) = (1/9)(4𝑥 − 𝑥^2), we have:

𝐸[𝑋^2] = ∫(𝑥^2(1/9)(4𝑥 − 𝑥^2))𝑑𝑥 = (1/9)∫(4𝑥^3 − 𝑥^4)𝑑𝑥 = (1/9)[(𝑥^4) − (1/5)𝑥^5]_0^3 = (1/9)[(3^4) − (1/5)(3^5)] = (1/9)(81 − 27) = 6

Therefore, 𝐸[𝑋^2] = 6.

To find the variance of 𝑋, we can use the formula:

var[𝑋] = 𝐸[𝑋^2] − (𝐸[𝑋])^2

We already know that 𝐸[𝑋^2] = 6 and 𝐸[𝑋] = 1.75 (as given in the question). Substituting these values gives:

var[𝑋] = 6 − (1.75)^2 = 6 − 3.0625 = 2.9375

Therefore, var[𝑋] = 2.9375.