Probability Question

Question

Let X be a random variable such that E[X] = 2, E[X^3 ] = 9 and E[(X – 2)^3 ] = 0. What is the

Var[X]?

Answer 1

Hello!

First and foremost, the equation for variance is: Var(X) = E[ X² ] - E[ X ]²

Our first step is to solve for E[ X² ] which we can do by expanding E[ (X-2)² ] which is E[ X³ - 6X² + 12X - 8 ]

Using the rule of expectations, namely that: E[ X + Y ] = E[X] + E[Y] and E[ cX ] = cE[X] we attain

E[ X³ - 6X² + 12X - 8 ] = E[ X³ ] -6 E[X²] + 12E[X] - 8

Furthermore, we know this is equal to 0.

E[ X³ ] - 6E[X²] + 12E[X] - 8 = 0.

We are given that E[X] = 2, and E[X³] = 9. What we can do is plug these values and use algebra to solve for E[X²]

9 - 6E[X²] + 24 - 8 = 0 ⇒ -6E[ X² ] = -25 ⇒ E[ X² ] = 25/6.

Now we can apply our formula for variance since we've solved for E[ X² ].

Var(X) = E[ X² ] - E[ X ]² = 25/6 - 2² = 25/6 - 4

Further simplification yields that this is equal to 1/6.

Hope this helps!