CH12 - Expected Values Continued

This lesson will continue the discussion of expected values, focusing on the expected value of sums and constants.

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Theorem To Remember

Theorem

If $x_1,x_2,\dots ,x_n$ are identically distributed discrete random variables, then:

$$E(X_1+X_2+\cdots +X_n)=nE(X_1)$$

and

$$E(a_1{X}_1+a_2{X}_2+\dots a_n{X}_n)=a_{1}E(X_1)+a_{2}E(X_2)+\cdots +a_{n}E(X_n)$$ $$E(aX+b)=aE(X)+b$$

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Examples

Example

Four students order noodles at a certain local restaurant. Their orders are placed independently. Each student is kmown to prefer Japanese pan noodles $40\%$ of the time. How many of them do we expect to order Japanese Pan Noodles?

Use X as an indicator variable and the theorem we learned to solve this problem

For each student:

$X_i$	$P_{X_i}(X_i)$, where $i=\{1,2,3,4\}$
0	.6
1	.4

n=4 students, all are iid

Let $A_j$ be the event that the $j_{th}$ student orders JPN

Let $X_j$ be the indicator:

$$\to E(X_j)=P(A_j)=.4$$

By the theorem

$$\begin{array}{rl}E(X)& =E(X_1+X_2+X_3+X_4)\\ & =4E(X_j)\\ & =4P(A_j)\\ & =4(.4)\\ & =4(.4)\\ & =1.6\end{array}$$

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Let’s revisit an older problem we love:

Example: Drawing cards without replacement until we get Ace of Hearts

We used a method in chapter 10 to find this by finding the sum of all the random variables. Let’s try some other methods.

Method 1

Let $A_j$ be the event that at least $j$ draws are required to get the ace of hearts.

$$\left[\begin{array}{c}{A}_{5}meansyoustilldraw5ormorecardstogetAceofHearts\\ \to donotgetaceofheartsonfirst5cards\end{array}\right]$$

then:

$$P(A_j)=1-P(A_j^c)=1-\frac{(j-1)}{52}$$

each draw is $\frac{1}{52}$ and there are $(j-1)$ of them.

Let $X_j$ be the indicator of $A_j$ $X_j\in \{0,1\}\leftarrow$ if $A_j$ occurs

Therefore:

$$E(X_j)=P(A_j)=1-\frac{(j-1)}{52}$$

for each $j=\{1,2,3,\dots ,52\}$

We have $X=X_1+\cdots +X_{52}$:

$$\begin{array}{rl}E(X)& =(X_1+\cdots +X_{52})\\ & =E(X_1)+\cdots +E(X_{52})\\ & =\sum _{j=1}^{52}\left(1-\frac{j-1}{52}\right)\\ & =52-\frac{1}{52}\sum _{j=1}^{52}(j-1)\\ & =52-\frac{1}{52}\left(\frac{51(52)}{2}\right)\\ & =51-\frac{51}{2}\\ & =26.5\end{array}$$