§ Linearity of expectation for sampling

# process 1
def val(c): return 1 + ord(c) - 'a'
def process(addx, addy):
    s = 0
    ndraws = 10
    for _ in range(ndraws):
        x = random.choice("abcde"),  # draw a random chit
        y = random.choice(x*5+"abcde") # draw a random chit, dependent on first random chit.
        if addx: s += val(x)
        if addy: x += val(y)
    return s
  • The key thing to remember is that the random variable does not care about the process, only about the value that is spit out by the process.
  • Thus we can read linearity of expectation as saying either (1) Simulate the full process in one go and accumulate the results as you go along process(True, True) or E[a+b], or (2) Simulate the process in parts, and add up the accumulated results from the partial simulations ( process(True, False) + process(False, True) or E[a] + E[b]).
  • In both cases , we are allowed to simulate the process fully! The only thing that differs is when we accumulate the answers.
  • This is in contrast to computing conditional probability, where the situation/process in which P(A|B) occurs is wildly different from P(A).
  • Linearity of expectation asks us to run the same process, just tally results differently.
  • It tells us that randomness allows the tallies to line up, whether we tally in two separate phases or in a single phase, which makes intuitive sense!

§ Linearity of expectation is purity

Suppose we write:
x = random.choice("abcde")
y = random.choice("abcde")
s =  val(x) + val(y)
  • If we ask for the expected value of s. It's going to be:
E[s] = E[val(x) + val(y)]
= E[val(x)] + E[val(y)]
= 2 E[val(x)]
  • The last inequality follows because x and y are two copies of the same random variable random.choice("abcde"), thus have the same expected value for val(x), val(y).
  • So, expecatation 'purifies' random computations.

§ "Deriving" equivalence for two processses using purity

  • First write down what process(True, False) + process(False, True) as:
def rhsI():
    sx = 0; sy = 0
    ndraws = 10
    for _ in range(ndraws):
        x = random.choice("abcde"),  # draw a random chit
        y = random.choice(x1*5+"abcde") # draw a random chit, dependent on first random chit.
        sx += val(x)

    for _ in range(ndraws):
        x = random.choice("abcde"),  # draw a random chit
        y = random.choice(x2*5+"abcde") # draw a random chit, dependent on first random chit.
        sy += val(y)
    return sx + sy
  • Next, we use the purity of random.choice (within expectation) to fuse the two loops:
def rhsII():
    sx = 0; sy = 0
    ndraws = 10

    # loop fusion is safe, because even though random.choice has a side effect, the order
    # of calling random.choice does not matter. It commutes with other random ops.
    for _ in range(ndraws):
        x1 = random.choice("abcde"),  # draw a random chit
        y1 = random.choice(x1*5+"abcde") # draw a random chit, dependent on first random chit.
        sx += val(x1)
        # loop fusion
        x2 = random.choice("abcde"),  # draw a random chit
        y2 = random.choice(x2*5+"abcde") # draw a random chit, dependent on first random chit.
        sy += val(y2)
    return sx + sy
  • Next, we use purity to set x1 = x2 and y1 = y2, since on expectation, their values are the same.
def rhsIII():
    sx = 0; sy = 0
    ndraws = 10

    # once again, expectation purifies randomness. So within the context of expecattion, we can
    # replace `x2` with `x1` with `x1`
    for _ in range(ndraws):
        x1 = random.choice("abcde"),  # draw a random chit
        y1 = random.choice(x1*5+"abcde") # draw a random chit, dependent on first random chit.
        sx += val(x1)
        # loop fusion
        x2 = x1
        y2 = y1
        sy += val(y2)
    return sx + sy
  • Finally, we cleanup the code to arrive at process(True, True):
def rhsIV():
    sx = 0; sy = 0
    ndraws = 10

    # once again, expectation purifies randomness. So within the context of expecattion, we can
    # replace `x2` with `x1` with `x1`
    for _ in range(ndraws):
        x1 = random.choice("abcde"),  # draw a random chit
        y1 = random.choice(x1*5+"abcde") # draw a random chit, dependent on first random chit.
        sx += val(x1)
        sy += val(y1)
    return sx + sy

§ For N=1N=1, the expected number of turns is 11.