Question

prove that E(x²) >> [E(X)]²?​

Answer 1

Answer:

I want to understand something about the derivation of Var(X)=E[X2]−(E[X])2

Variance is defined as the expected squared difference between a random variable and the mean (expected value): Var(X)=E[(X−μ)2]

Then:

Var(X)=E[(X−μ)2]

Var(X)=E[(X−E[X])2]

Var(X)=E[(X−E[X])(X−E[X])]

Var(X)=E[X2−2XE[X]+(E[X])2]

Var(X)=E[X2]−2E[XE[X]]+E[(E[X])2]

Var(X)=E[X2]−2E[E[X]E[X]]+E[(E[X])2]

Var(X)=E[X2]−2(E[X])2+(E[X])2

Var(X)=E[X2]−(E[X])2

What I don't quite understand is the steps that get us from E[XE[X]] to E[E[X]E[X]] to (E[X])2, also E[(E[X])2] to (E[X])2.

While I'm sure these jumps are intuitive and obvious I would still like to understand how we can (more formally) make these jumps / consider them mathematically equivalent.

Please mark me as brainliest