# 24.5 - More Examples

24.5 - More Examples

On this page, we'll just take a look at a few examples that use the material and methods we learned about in this lesson.

## Example 24-4

If $$X_1,X_2,\ldots, X_n$$ are a random sample from a population with mean $$\mu$$ and variance $$\sigma^2$$, then what is:

$$E[(X_i-\mu)(X_j-\mu)]$$

for $$i\ne j$$, $$i=1, 2, \ldots, n$$?

#### 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:

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

## Example 24-5

Let $$X_1, X_2, X_3$$ be a random sample of size $$n=3$$ from a distribution with the geometric probability mass function:

$$f(x)=\left(\dfrac{3}{4}\right) \left(\dfrac{1}{4}\right)^{x-1}$$

for $$x=1, 2, 3, \ldots$$. What is $$P(\max X_i\le 2)$$?

#### 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:

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

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