Question

Suppose that \(p(x)=\frac{1}{5}, x=1,2,3,4,5\), zero elsewhere, is the pmf of the discrete-type random variable \(X\). Compute \(E(X)\) and \(E\left(X^{2}\right)\). Use these two results to find \(E\left[(X+2)^{2}\right]\) by writing \((X+2)^{2}=X^{2}+4 X+4\).

Short answer

The expected values are as follows: \(E(X) = 3\), \(E(X^2) = 11\) and \(E((X+2)^2) = 26\).

Step-by-step solution

Calculating Expected Value \(E(X)\)

The expected value of a discrete random variable \(X\) is given by \(E(X) = \sum x*p(x)\). In this case, that is equal to \(\frac{1}{5}* 1 + \frac{1}{5}* 2 + \frac{1}{5}* 3 + \frac{1}{5}* 4+ \frac{1}{5}* 5 = 3\).

Calculating Expected Value of \(E(X^2)\)

The expected value of \(X^2\) can be computed by \(\sum x^2*p(x)\). Defining \(X^2\) as \(Y\) and using the given pmf, we get \(\frac{1}{5}*1^2 + \frac{1}{5}*2^2 + \frac{1}{5}*3^2 + \frac{1}{5}*4^2 + \frac{1}{5}* 5^2 = 11\).

Calculating Expected Value of \((X+2)^2\)

Now, to calculate \(E((X+2)^2)\), substitute the equation \((X+2)^2= X^2+4X+4 \) into expected value formula: \(E((X+2)^2) = E(X^2 + 4X + 4) = E(X^2) + 4*E(X) + 4 = 11 + 4*3 + 4 = 26\).