Expectation formula [t) Prove that E[X2] |g E[X] for any random variable X. w Gi
ID: 2989191 • Letter: E
Question
Expectation formula [t) Prove that E[X2] |g E[X] for any random variable X. w Give an example that E[X2] > E[X] Give examples for the following cases or explain why such case does not exist X1 and X2 two identically distributed independent rv. X1 and X2 two identically distributed dependent r, v. X1 and X2 non-identical but independent rv. X1 and X2 non-identical and also dependent rv. (10%) Prove that for the Poisson distribution with Px(n) = xn/n! e"x:n = 0, 1,2, 3, .... Expected value E[X] = x Variance E[(X-E[X])2] = x.Explanation / Answer
5.)
a.) We know Var(X) >= 0
Hence,
E(X - E(X))2 >= 0
= E(X2 + E2(X) - 2XE(X)) >= 0
= E(X2) + E(E2(X)) - 2E(XE(X)) >= 0
= E(X2) + E2(X) - 2E2(X) >= 0
= E(X2) - E2(X) >= 0
Hence proved
b.) Example
X is a random variable of no. of head appearing in two toss
Hence,
E(X) = 1P(1)+2P(2) = 1/2 + 2(1/4) = 1
E2(X) = 1
E(X2) = 1P(1)+22P(2) = 1/2 + 4(1/4) = 1.5
E(X2) = 1.5 > 1 = E2(X)
6.)
a.) X1 and X2 are iid random variable with normal distribution N(0,t)
b.) X1 and X2 are identical random variable with normal distribution N(0,t) with Cov(X1 , X2) > 0
c.) X1 and X2 are independent random variable with
X1 ~normal distribution N(0,t) and
X2 ~exp(d)
d.) X1 and X2 are dependent random variable [i.e. cov(X1,X2)>0] with
X1 ~normal distribution N(0,t) and
X2 ~exp(d)
7.)
a.)
E(X) = 1P(1) + 2P(2) + .... = e-x[1*x + 2*x2/2! + 3*x3/3! ..... ]
= e-x[x*d(ex)/dx ]
= e-x[x*ex] = x
b.)
E(X2) = 1P(1) + 22P(2) + .... = e-x[1*x + 22*x2/2! + 32*x3/3! ..... ]
= e-xd[x*d(ex)/dx] / dx
= e-x*x*d[xex]/dx
= e-x*x*[ex + xex]
= x + x2
Var(X) = E(X2) - E2(X) = x + x2 - x2 = x
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