Key Concepts and Given Information

We are given a linear transformation $X = aZ + b$, where $Z$ is a standard normal random variable ($Z \sim N(0, 1)$). This means $E[Z] = 0$ and $Var(Z) = E[Z^2] = 1$. We need to find the constants $a$ and $b$ using the provided expectation conditions.

• $E[X] = 1$
• $E[(X – E[X])Z] = –2$
• $E[(X – E[X])^2] = 4$, which is equivalent to $Var(X) = 4$.

Solving for Constant b

From the condition $E[X] = 1$:

$ E[aZ + b] = 1 $

Using the linearity property of expectation:

$ aE[Z] + E[b] = 1 $

Since $E[Z] = 0$ and $E[b] = b$:

$ a(0) + b = 1 $

$ \implies b = 1 $

Solving for Constant a

From the condition $Var(X) = 4$:

$ Var(aZ + b) = 4 $

Using the property $Var(aZ + b) = a^2 Var(Z)$:

$ a^2 Var(Z) = 4 $

Since $Var(Z) = 1$:

$ a^2 (1) = 4 $

$ \implies a^2 = 4 $

This implies $a = 2$ or $a = -2$. We use the remaining condition to find the specific value of $a$.

Using the Second Expectation Condition

From the condition $E[(X – E[X])Z] = –2$:

Substitute $X = aZ + b$ and $E[X] = 1$. Since we found $b = 1$, $X = aZ + 1$ and $E[X] = 1$.

$ E[((aZ + 1) – 1)Z] = –2 $

Simplify the expression inside the expectation:

$ E[(aZ)Z] = –2 $

$ E[aZ^2] = –2 $

Using the linearity of expectation:

$ aE[Z^2] = –2 $

Since $E[Z^2] = Var(Z) = 1$:

$ a(1) = –2 $

$ \implies a = -2 $

Conclusion

We found $b = 1$ and $a = -2$. Therefore, the values are $a = -2$ and $b = 1$.