Solution

We can calculate the mean and variance of \(Y\) in three different ways.

1. By recognizing that \(Y\) is a binomial random variable with \(n=5\) and \(p=\frac{1}{2}\), we can use what know about the mean and variance of a binomial random variable, namely that the mean of \(Y\) is:

   \(E(Y)=np=5(\frac{1}{2})=\frac{5}{2}\)

   and the variance of \(Y\) is:

   \(Var(Y)=np(1-p)=5(\frac{1}{2})(\frac{1}{2})=\frac{5}{4}\)

   Since sums of independent random variables are not always going to be binomial, this approach won't always work, of course. It would be good to have alternative methods in hand!

2. We could use the linear operator property of expectation. Before doing so, it would be helpful to note that the mean of \(X_1\) is:

   \(E(X_1)=np=3(\frac{1}{2})=\frac{3}{2}\)

   and the mean of \(X_2\) is:

   \(E(X_2)=np=2(\frac{1}{2})=1\)

   Now, using the property, we get that the mean of \(Y\) is (thankfully) again \(\frac{5}{2}\):

   \(E(Y)=E(X_1+X_2)=E(X_1)+E(X_2)=\dfrac{3}{2}+1=\dfrac{5}{2}\)

   Recall that the second equality comes from the linear operator property of expectation. Now, using the linear operator property of expectation to find the variance of \(Y\) takes a bit more work. First, we should note that the variance of \(X_1\) is:

   \(Var(X_1)=np(1-p)=3(\frac{1}{2})(\frac{1}{2})=\frac{3}{4}\)

   and the variance of \(X_2\) is:

   \(Var(X_2)=np(1-p)=2(\frac{1}{2})(\frac{1}{2})=\frac{1}{2}\)

   Now, we can (thankfully) show again that the variance of \(Y\) is \(\frac{5}{4}\):

   Okay, as if two methods aren't enough, we still have one more method we could use.

3. We could use the independence of the two random variables \(X_1\) and \(X_2\), in conjunction with the definition of expected value of \(Y\) as we know it. First, using the binomial formula, note that we can present the probability mass function of \(X_1\) in tabular form as:

   \(x_1\)	0	1	2	3
   \(f(x_1)\)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)

   And, we can present the probability mass function of \(X_2\) in tabular form as well:

   \(x_2\)	0	1	2
   \(f(x_2)\)	\(\frac{1}{4}\)	\(\frac{2}{4}\)	\(\frac{1}{4}\)

   Now, recall that if \(X_1\) and \(X_2\) are independent random variables, then:

   \(f(x_1,x_2)=f(x_1)\cdot f(x_2)\)

   We can use this result to help determine \(g(y)\), the probability mass function of \(Y\). First note that, since \(Y\) is the sum of \(X_1\) and \(X_2\), the support of \(Y\) is {0, 1, 2, 3, 4 and 5}. Now, by brute force, we get:

   \(g(0)=P(Y=0)=P(X_1=0,X_2=0)=f(0,0)=f_{X_1}(0) \cdot f_{X_2}(0)=\dfrac{1}{8} \cdot \dfrac{1}{4}=\dfrac{1}{32}\)

   The second equality comes from the fact that the only way that \(Y\) can equal 0 is if \(X_1=0\) and \(X_2=0\), and the fourth equality comes from the independence of \(X_1\)and \(X_2\). We can make a similar calculation to find the probability that \(Y=1\):

   \(g(1)=P(X_1=0,X_2=1)+P(X_1=1,X_2=0)=f_{X_1}(0) \cdot f_{X_2}(1)+f_{X_1}(1) \cdot f_{X_2}(0)=\dfrac{1}{8} \cdot \dfrac{2}{4}+\dfrac{3}{8} \cdot \dfrac{1}{4}=\dfrac{5}{32}\)

   The first equality comes from the fact that there are two (mutually exclusive) ways that \(Y\) can equal 1, namely if \(X_1=0\) and \(X_2=1\) or if \(X_1=1\) and \(X_2=0\). The second equality comes from the independence of \(X_1\) and \(X_2\). We can make similar calculations to find \(g(2), g(3), g(4)\), and \(g(5)\). Once we've done that, we can present the p.m.f. of \(Y\) in tabular form as:

   \(y = x_1 + x_2\)	0	1	2	3	4	5
   \(g(y)\)	\(\frac{1}{32}\)	\(\frac{5}{32}\)	\(\frac{10}{32}\)	\(\frac{10}{32}\)	\(\frac{5}{32}\)	\(\frac{1}{32}\)

   Then, it is a straightforward calculation to use the definition of the expected value of a discrete random variable to determine that (again!) the expected value of \(Y\) is \(\frac{5}{2}\): 

   \(E(Y)=0(\frac{1}{32})+1(\frac{5}{32})+2(\frac{10}{32})+\cdots+5(\frac{1}{32})=\frac{80}{32}=\frac{5}{2}\)

   The variance of \(Y\) can be calculated similarly. (Do you want to calculate it one more time?!)

   The following summarizes the method we've used here in calculating the expected value of \(Y\):

   \begin{align} E(Y)=E(X_1+X_2) &= \sum\limits_{x_1 \in S_1}\sum\limits_{x_2 \in S_2} (x_1+x_2)f(x_1,x_2)\\ &= \sum\limits_{x_1 \in S_1}\sum\limits_{x_2 \in S_2} (x_1+x_2)f(x_1) f(x_2)\\ &= \sum\limits_{y \in S} yg(y)\\ \end{align}

   The first equality comes, of course, from the definition of \(Y\). The second equality comes from the definition of the expectation of a function of discrete random variables. The third equality comes from the independence of the random variables \(X_1\) and \(X_2\). And, the fourth equality comes from the definition of the expected value of \(Y\), as well as the fact that \(g(y)\) can be determined by summing the appropriate joint probabilities of \(X_1\) and \(X_2\).

Solution

We'll use the fact that the expectation of the product is the product of the expectations:

Solution

The mean of the sum is:

\(E(X_1+X_2)=E(X_1)+E(X_2)=2+3=5\)

and the variance of the sum is:

\(Var(X_1+X_2)=(1)^2Var(X_1)+(1)^2Var(X_2)=4+5=9\)

Solution

The mean of the difference is:

\(E(X_1-X_2)=E(X_1)-E(X_2)=2-3=-1\)

and the variance of the difference is:

\(Var(X_1-X_2)=Var(X_1+(-1)X_2)=(1)^2Var(X_1)+(-1)^2Var(X_2)=4+5=9\)

That is, the variance of the difference in the two random variables is the same as the variance of the sum of the two random variables.

Solution

The mean of the linear combination is:

\(E(3X_1+4X_2)=3E(X_1)+4E(X_2)=3(2)+4(3)=18\)

and the variance of the linear combination is:

\(Var(3X_1+4X_2)=(3)^2Var(X_1)+(4)^2Var(X_2)=9(4)+16(5)=116\)

Solution

The fact that \(X_1,X_2,\ldots, X_n\) constitute a random sample tells us that (1) \(X_i\) is independent of \(X_j\), for all \(i\ne j\), and (2) the \(X_i\) are identically distributed. Now, we know from our previous work that if \(X_i\) is independent of \(X_j\), for \(i\ne j\), then the covariance between \(X_i\) is independent of \(X_j\) is 0. That is:

\(E[(X_i-\mu)(X_j-\mu)]=Cov(X_i,X_j)=0\)

Solution

The only way that the maximum of the \(X_i\) will be less than or equal to 2 is if all of the \(X_i\) are less than or equal to 2. That is:

\(P(\max X_i\leq 2)=P(X_1\leq 2,X_2\leq 2,X_3\leq 2)\)

Now, because \(X_1,X_2,X_3\) are a random sample, we know that (1) \(X_i\) is independent of \(X_j\), for all \(i\ne j\), and (2) the \(X_i\) are identically distributed. Therefore:

\(P(\max X_i\leq 2)=P(X_1\leq 2)P(X_2\leq 2)P(X_3\leq 2)=[P(X_1\leq 2)]^3\)

The first equality comes from the independence of the \(X_i\), and the second equality comes from the fact that the \(X_i\) are identically distributed. Now, the probability that \(X_1\) is less than or equal to 2 is:

\(P(X\leq 2)=P(X=1)+P(X=2)=\left(\dfrac{3}{4}\right) \left(\dfrac{1}{4}\right)^{1-1}+\left(\dfrac{3}{4}\right) \left(\dfrac{1}{4}\right)^{2-1}=\dfrac{3}{4}+\dfrac{3}{16}=\dfrac{15}{16}\)

Therefore, the probability that the maximum of the \(X_i\) is less than or equal to 2 is:

\(P(\max X_i\leq 2)=[P(X_1\leq 2)]^3=\left(\dfrac{15}{16}\right)^3=0.824\)