Problem 33 - A First Course in Probability

Question

If \(E[X]=1\) and \(\operatorname{Var}(X)=5,\) find (a) \(E\left[(2+X)^{2}\right]\) (b) \(\operatorname{Var}(4+3 X)\)

Step-by-Step Solution

Answer

(a) \(E\left[(2+X)^{2}\right] = 14\) (b) \(\operatorname{Var}(4+3X) = 45\)

Step 1: (a) Calculate E[((2+X)^2)] #

We start by expanding the expression within the expected value operator: \[ (2+X)^2 = 4 + 4X + X^2 \] Now, we can use the linearity of expected value, that is: \[ E[a + b\,Y] = a + b\,E[Y] \] Applying this property to our problem: \[ E[(2+X)^2] = E[4+4X+X^2] = E[4] + 4E[X] + E[X^2] \] We know that \(E[X]=1\). To find \(E[X^2]\), we will use the relation between variance and second moment: \[ \operatorname{Var}(X) = E[X^2] - (E[X])^2 \Rightarrow E[X^2] = \operatorname{Var}(X) + (E[X])^2 \] Now, we can substitute the given values and find \(E[X^2]\): \[ E[X^2] = 5 + 1^2 = 6 \] Finally, we can find \(E[(2+X)^2]\): \[ E[(2+X)^2] = E[4] + 4E[X] + E[X^2] = 4 + 4(1) + 6 = 14 \]

Step 2: (b) Calculate Var(4+3X) #

We will use the following property of variance, for a transformation of a random variable \(Y = a + bX\): \[ \operatorname{Var}(a + b\,X) = b^2 \operatorname{Var}(X) \] In our case, we have \(a = 4\) and \(b = 3\), so we can find \(\operatorname{Var}(4+3X)\): \[ \operatorname{Var}(4+3X) = 3^2 \operatorname{Var}(X) = 9(5) = 45 \] So the final answers are: (a) \(E\left[(2+X)^2\right] = 14\). (b) \(\operatorname{Var}(4+3X) = 45\).