## Monday, June 2, 2014

### Section 2.5 - Sum Expected Values

What is the expected value of the sum $$X + Y$$ given $$X$$ and $$Y$$ are (not necessarily independent) random variables?

From the definition of expected value, $E(X + Y) = \sum_x\sum_y (x + y)\,p(X = x \wedge Y = y)$ Evaluating this sum as given requires work proportional to $$n^2$$; can it be evaluated more efficiently? Because $$X$$ and $$Y$$ aren’t necessarily independent, the probability $$p(X = x \wedge Y = y)$$ can’t be simplified in the same way it was when computing the expected value of random-value products.? Are there some other tricks that can simplify evaluation?

## Friday, May 30, 2014

### Section 2.5 - Product Expected Value

Given independent random variables $$X$$ and $$Y$$,? what is the expected value of the product $$XY$$?

From the definition of expected vlaue, $E(XY) = \sum_x\sum_y xy\,p(XY = xy)$ Because $$X$$ and $$Y$$ are independent, the probability of their product is the product of their probabilities $p(XY = xy) = p_X(X = x)\,p_Y(Y = y)$ and the expected value becomes $E(XY) = \sum_x\sum_y xy\,p_X(X = x)\,p_Y(Y = y)$ where $$p_X$$ is the probability distribution for $$X$$ and similarly for $$p_Y$$.?

Evaluating the expected value as given takes work proportional to $$n^2$$. Can the sum be evaluated with less work?

## Sunday, June 19, 2011

### Section 2.5 - Random Variables, Mean and the Expected Value

A random variable maps outcomes from a sample space to numbers, making the outcomes easier to manipulate. For example, there isn’t much to be done with outcomes from the coin-flip sample space { H, T }, but mapping outcomes to integers with the random variable X = { (H, 0), (T,1) } allows mathematical manipulations. The mapping used depends on what’s being done. X's previous definition might be useful when counting the number of tails, while the definition X = { (H, 1), (T, -1) } is more suited for keeping track of wins and losses.

Given a random variable X over a sample space, the probability that Xequals a particular value k is the sum of the probabilities of the outcomes that X maps to k: $Pr(X = k) = \sum_{X(o) = k} Pr(o)$ For example, let the random variable X map the outcome i of a single die roll to i. The probability that X is even is $Pr(even(X)) = Pr(2)+ Pr(4) + Pr(6) = 1/6 + 1/6 + 1/6 = 1/2$Computing the mean (or average or expected value) of a random value is a common mathematical manipulation. A random variable’s mean is the sum of the values in the range, with each value weighted by its probability: $\sum_{k\in range(X)} kPr(X = k)$ The mean is often represented by $$\mu$$. To continue the die-roll example, X's mean is
$\begin{eqnarray} & & 1Pr(X = 1) + 2Pr(X = 2) + \cdots + 6Pr(X = 6) \\ &=& 1\cdot1/6 + 2\cdot1/6 + \cdots + 6\cdot1/6 \\ &=& (1 + 2 + \cdots + 6)1/6 \\ &=& 21\cdot1/6 \\ &=& 3.5 \end{eqnarray}$A random variable’s mean is not necessarily a member of its range.

## Friday, June 17, 2011

### Exercise 2.2-16

How many distinct three-letter combinations (“words”) can you make from the letters of Mississippi?

## Tuesday, May 31, 2011

### Exercise 2.2-3

What is the probability of cutting two cards of the same rank from a well-shuffled deck?

## Monday, May 30, 2011

### Section 2.4 - The Binomial Distribution—Bernoulli Trials

A Bernoulli trial produces an outcome which satisfies a criterion with probability $$p$$, and so fails to satisfy the criterion with compliment probability $$q = 1 - p$$. What is the probability that $$n$$ successive, independent Bernoulli trials results in exactly $$k$$ satisfactory outcomes?

The compound outcome is an arrangement of $$k$$ satisfactory outcomes and $$n - k$$ unsatisfactory outcomes. Because the Bernoulli-trial outcomes are independent, the outcome probabilities are multiplied to get the combined-outcome probability, and because multiplication is commutative, any individual compound outcome has probability $$p^kq^{n-k}$$. There are $$C(n, k)$$ such compound outcomes, and the probability that $$n$$ successive, independent Bernoulli trials results in exactly $$0 \leq k \leq n$$ satisfactory outcomes when satisfactory outcomes have probability $$p$$ is $b(k; n, p) = C(n, k)p^kq^{n - k}$ The compound-outcome sample space is not uniform. Even if $$p = 1/2$$,? $$C(n, k)$$ varies with $$k$$; the variation becomes more complex when $$p\neq q$$.

## Sunday, May 29, 2011

### Section 2.3 - Combinations

Given $$n$$ distinct items, there are $$P(n, k)$$ permutations of $$0 \leq k \leq n$$ items. Each permutation of $$k$$ items can be re-arranged $$k!$$ ways. If item order doesn’t matter, there are
$\frac{P(n, k)}{k!} = \frac{n!}{k!(n-k)!} = C(n, k)$
orderless permutations, or combinations. Sanity check: $$C(n, n) = n!/(n!(n - n)!) = 1/0! = 1$$.

Combinations are fun to play with. First, $$C(n, k) = C(n, n - k)$$.? Then
$\begin{eqnarray} C(n, k + 1) & = & \frac{n!}{(k + 1)!(n - k - 1)!} \\ & = & \frac{n!(n - k)}{(k + 1)k!(n - k)(n - k - 1)!} \\ & = & \frac{n - k}{k + 1}C(n, k) \end{eqnarray}$
$$C(n, k + 1)$$ is largest when $$(n - k)/(k + 1)$$ is closest to 1, which occurs when $$k$$ is closest to $$(n - 1)/2$$. Similarly
$\begin{eqnarray} C(n + 1, k) & = & \frac{(n + 1)!}{k!(n + 1 - k)!} \\ & = & \frac{(n + 1)n!}{k!(n + 1 - k)(n - k)!} \\ & = & \frac{n + 1}{n + 1 - k}C(n, k) \end{eqnarray}$