Let X = aZ + b, where Z is a standard normal random variable, and a, b are two unknown constants. It is given that

E[X] = 1, E[(X − E[X])Z] = −2, E[(X − E[X])2 ] = 4,

where E[X] denotes the expectation of random variable X. The values of a, b are:

A.

a = −2, b = 1

B.

a = 2, b = −1

C.

a = −2, b = −1

D.

a = 1, b = 1

Solution:

We are given a random variable:

X=aZ+bX = aZ + b

where Z∼N(0,1)Z \sim \mathcal{N}(0,1) (standard normal), and constants aa and bb are to be found.

We're also given:

  1. E[X]=1E[X] = 1

  2. E[(X−E[X])Z]=−2E[(X - E[X])Z] = -2

  3. E[(X−E[X])2]=4E[(X - E[X])^2] = 4

Step 1: Compute E[X]E[X]

Since Z∼N(0,1)Z \sim \mathcal{N}(0,1), we have E[Z]=0E[Z] = 0. So:

E[X]=E[aZ+b]=a⋅E[Z]+b=0+b=bE[X] = E[aZ + b] = a \cdot E[Z] + b = 0 + b = b

Given: E[X]=1⇒b=1E[X] = 1 \Rightarrow b = 1

Step 2: Compute E[(X−E[X])Z]E[(X - E[X])Z]

We already have:

X=aZ+b,E[X]=b=1⇒X−E[X]=aZ+b−1=aZX = aZ + b,\quad E[X] = b = 1 \Rightarrow X - E[X] = aZ + b - 1 = aZ

So:

E[(X−E[X])Z]=E[aZ⋅Z]=a⋅E[Z2]=a⋅Var(Z)=a⋅1=aE[(X - E[X])Z] = E[aZ \cdot Z] = a \cdot E[Z^2] = a \cdot \text{Var}(Z) = a \cdot 1 = a

Given: E[(X−E[X])Z]=−2⇒a=−2E[(X - E[X])Z] = -2 \Rightarrow a = -2

Step 3: Compute E[(X−E[X])2]E[(X - E[X])^2]

X−E[X]=aZ=−2Z⇒(X−E[X])2=a2Z2=4Z2X - E[X] = aZ = -2Z \Rightarrow (X - E[X])^2 = a^2 Z^2 = 4Z^2E[(X−E[X])2]=4⋅E[Z2]=4⋅1=4(satisfied)E[(X - E[X])^2] = 4 \cdot E[Z^2] = 4 \cdot 1 = 4 \quad \text{(satisfied)}

✅ Final Answer: (A)

a=−2,b=1a = -2, b = 1