Calculate the expectation of a discrete random variable, using the example of tossing four coins.

Question

ScanSolve AI Tutor · Accepted Answer

Here's a breakdown of the concepts from your notes:

Expectation (Mean) of a Discrete Random Variable: The expectation, or mean, of a discrete random variable $X$ is denoted by $E[X]$ or $\mu$ . It represents the average value of the random variable over many trials. The formula for the expectation is:
$E[X] = \sum_{x} x P(X=x)$
where $x$ represents each possible value of the random variable, and $P(X=x)$ is the probability of $X$ taking that value.

Example: Tossing Four Coins
The notes mention "tossing three coins" but the probabilities are given out of 16, which corresponds to tossing four coins ( $2^4 = 16$ possible outcomes). Let $X$ be the number of heads when tossing four fair coins. The possible values for $X$ are $0, 1, 2, 3, 4$ , with the following probabilities:
$P(X=0) = 1/16$
$P(X=1) = 4/16$
$P(X=2) = 6/16$
$P(X=3) = 4/16$
$P(X=4) = 1/16$

Substitute the values into the expectation formula.

$$
E[X] = 0 \left(\frac{1}{16}\right) + 1 \left(\frac{4}{16}\right) + 2 \left(\frac{6}{16}\right) + 3 \left(\frac{4}{16}\right) + 4 \left(\frac{1}{16}\right)
$$

Multiply the terms.

$$
E[X] = \frac{0}{16} + \frac{4}{16} + \frac{12}{16} + \frac{12}{16} + \frac{4}{16}
$$

Sum the numerators.

$$
E[X] = \frac{0+4+12+12+4}{16} = \frac{32}{16}
$$

Simplify the fraction.
$$ E[X] = \boxed{2} $$
- Variance of a Discrete Random Variable: The variance of a discrete random variable $X$ , denoted by $\sigma^2$ or $Var(X)$ , measures how far the values of the random variable are spread out from the mean. The formula for variance is:
  $\sigma^2 = E[(X-\mu)^2]$
  This can be expanded and simplified to a more computationally convenient form:
  $\sigma^2 = E[X^2] - (E[X])^2$
  where $E[X^2] = \sum_{x} x^2 P(X=x)$ .
  
  Example: Calculating $E[X^2]$ for Tossing Four Coins
  Using the same example of tossing four coins, we first calculate $E[X^2]$ .

Substitute the values into the $E[X^2]$ formula.

$$
E[X^2] = 0^2 \left(\frac{1}{16}\right) + 1^2 \left(\frac{4}{16}\right) + 2^2 \left(\frac{6}{16}\right) + 3^2 \left(\frac{4}{16}\right) + 4^2 \left(\frac{1}{16}\right)
$$

Square the $x$ values and multiply.

$$
E[X^2] = 0 \left(\frac{1}{16}\right) + 1 \left(\frac{4}{16}\right) + 4 \left(\frac{6}{16}\right) + 9 \left(\frac{4}{16}\right) + 16 \left(\frac{1}{16}\right)
$$
$$
E[X^2] = \frac{0}{16} + \frac{4}{16} + \frac{24}{16} + \frac{36}{16} + \frac{16}{16}
$$

Sum the numerators.

$$
E[X^2] = \frac{0+4+24+36+16}{16} = \frac{80}{16}
$$

Simplify the fraction.

$$
E[X^2] = \boxed{5}
$$

**Calculating the Variance**  
Now we can calculate the variance using $E[X] = 2$ and $E[X^2] = 5$.

Use the variance formula.
$$ \sigma^2 = E[X^2] - (E[X])^2 $$
Substitute the calculated values.
$$ \sigma^2 = 5 - (2)^2 $$

Simplify.

$$
\sigma^2 = 5 - 4
$$
$$
\sigma^2 = \boxed{1}
$$

Calculate the expectation of a discrete random variable, using the example of tossing four coins.

Answer

Step-by-step solution

Substitute the values into the expectation formula.

Multiply the terms.

Sum the numerators.

Simplify the fraction.

Substitute the values into the $E[X^2]$ formula.

Square the $x$ values and multiply.

Sum the numerators.

Simplify the fraction.

Use the variance formula.

Substitute the calculated values.

Simplify.

Still stuck on this one?

More Mathematics Questions

Calculate the expectation of a discrete random variable, using the example of tossing four coins.

Answer

Step-by-step solution

Substitute the values into the expectation formula.

Multiply the terms.

Sum the numerators.

Simplify the fraction.

Substitute the values into the E[X2]E[X^2]E[X2] formula.

Square the xxx values and multiply.

Sum the numerators.

Simplify the fraction.

Use the variance formula.

Substitute the calculated values.

Simplify.

Still stuck on this one?

More Mathematics Questions

Substitute the values into the $E[X^2]$ formula.

Square the $x$ values and multiply.