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EECS 501 CONDITIONAL PROBABILITY Fall 2001 Three There are 3 cards: red/red; red/black; black/black. Card The cards are shuffled and one chosen at random. Monte The top of the card is red. Pr[bottom is red]=?
3 possible lines of reasoning to solve this problem:
- Bottom is red only if chose red/red card→ P r = 1/3.
- Not black/black, so either red/black or red/red→ P r = 1/2.
- 5 hidden sides: 2 red and 3 black→ P r = 2/5. Which is correct? They are ALL wrong! In fact, P r = 2/3.
DEF: P r[A|B] = P r[event A occurs, GIVEN THAT event B occurrED].
- Either A occurs or A doesn’t occur, even if B occurred.
- Their relative probabilities (ratio) shouldn’t change, after restriction to B occurring (A ∩ B and A′^ ∩ B). Want: P r[A|B] + P r[A′|B] = 1 and (^) P rP r[[AA′||BB]] = (^) P rP r[[AA′∩∩BB]]. Know: P r[A ∩ B] + P r[A′^ ∩ B] = P r[B]. So just divide this by P r[B].
THM: P r[A|B] = P r[A ∩ B]/P r[B] = P r[A ∩ B]/(P r[A ∩ B] + P r[A′^ ∩ B]). NOTE: Forms: P r[A|B] = (^) x+xy and P r[A′|B] = (^) x+yy. Ratio x/y, add to one.
EX: P r[Bottom red|T op red] = P r[T op AN D Bottom red P r[T op red] ] = 11 //^32 = 2/3.
OR: P r[BR|T R] = (^) P r[RR]P r[T R|RR]+P r[RBP r]P r[RR[T R]|RB]+P r[BB]P r[T R|BB] = (^) (1/3)(1)+(1/3)(1^1 /^3 /2)+(1/3)(0) = 2/3 (try drawing a 3-branch tree).
Why? BB eliminated; 1 red face up; 2 of 3 remaining faces are red!
Bayes’s P r[A|B] = P r P r[A[∩BB] ] = P r[B|A] (^) P rP r[[AB]] = (^) P r[B|A]P rP r[[AB]+|AP r]P r[B[A|A]′]P r[A′].
Why? Suppose we are given P r[A] (a priori prob. of A occurring). Now we observe B occurs. How does this change prob. of A? i.e.: Compute a posteriori prob. P r[A|B] of A occurring, given B.
EX: Coin A has Pr[heads]=1/3; Coin B has Pr[heads]=3/4. Choose coin at random, flip it, get heads. Compute Pr[coin A]. Note: Without observing the flip result, Pr[coin A]=1/2 (a priori). But: P r[A|H] = (^) P r[H|A]P rP r[[HA|]+A]P rP r[[HA|]B]P r[B] = (^) (1/3)(1(1//2)+(33)(1//2)4)(1/2) = 134 < 12.
Why? Observed heads→more likely chose coin more likely to land heads.
EECS 501 CONTINUITY OF PROBABILITY Fall 2001
DEF: Increasing sequence of sets A 1 ⊂ A 2 ⊂ A 3 ⊂... (^) nlim→∞ An = ∪∞ n=1An. DEF: Decreasing sequence of sets A 1 ⊃ A 2 ⊃ A 3 ⊃... (^) nlim→∞ An = ∩∞ n=1An. Thm: P r[ (^) nlim→∞ An] = (^) nlim→∞ P r[An] for either increasing or decreasing sets. Note: We can interchange ”limit” and the function ”P r”; P r is continuous.
Proof: For the increasing sequence {An}, let Bn = An − An− 1 , A 0 = φ. P r[An] = P r[∪ni=1Ai] = P r[∪ni=1Bi] =
∑n i=1 P r[Bi]^ (Bi^ ∩^ Bj^ =^ φ). lim: (^) nlim→∞ P r[An] =
i=1 P r[Bi] =^ P r[∪
∞ i=1Bi] =^ P r[^
lim n→∞ An] using the third axiom for countably infinite union of disjoint Bi.
- The proof for a decreasing sequence of sets {An} is similar.
Note: If we start with disjoint {Bi} and define An = ∪ni=1Bi, and we suppose that continuity of probability is true, we can use this argument to derive the third axiom! Historically, this is the way Kolmogorov did it in 1933.
DEF: For any sequence of sets {An}, we can define the limsup and liminf: limsup n→∞ An^ =^
lim n→∞ ∪
∞ i=n Ai;^
liminf n→∞ An^ =^
lim n→∞ ∩
∞ i=n Ai. Apply continuity of probability using limsup and liminf, not lim.
- Property: ( limsup n→∞ An)′^ = liminf n→∞ A′ n; ( liminf n→∞ An)′^ = limsup n→∞ A′ n.
EX1: Spin a wheel of fortune. Compute P r[{ 12 }] using cont. of probability.
P r[{ 12 }] = P r[ (^) nlim→∞ ( 12 − (^) n^1 , 12 + (^1) n )] = (^) nlim→∞ P r[( 12 − (^1) n , 12 + (^) n^1 )] = (^) nlim→∞ n^2 = 0
since An = ( 12 − (^1) n , 12 + (^) n^1 ) is a decreasing sequence of sets. We already knew this, but now we can use this simpler derivation.
EX2: Spin a wheel. Compute Pr[decimal expansion WON’T contain a 6]. Let A = {x : 0 ≤ x < 1 } and the decimal expansion of x has no 6. A = [0, 1) − [0. 6 , 0 .7) − ∪^9 n= n 6 =
[0.n 6 , 0 .n7) − ∪^9 i= i 6 =
∪^9 j= j 6 =
[0.ij 6 , 0 .ij7) −... An+1 = An − Bn = An − ∪^9 i 1 = i 1 6 =
· · · ∪^9 in= in 6 =
[0.i 1... in 6 , 0 .i 1... in7). Bn ⊂ An → P r[An − Bn] = P r[An] − P r[Bn]. Decreasing sequence.
Cont. of prob.→ P r[A] = (^) nlim→∞ P r[An] = (^) nlim→∞ (1 −
∑n− 1 i=0 P r[Bi]) = (^) nlim→∞ (1− 0. 1 −9(0.1)^2 − 92 (0.1)^3 −.. .− 9 n−^1 (0.1)n) = (^) nlim→∞ (0.9)n^ = 0. Note: Heuristically, Pr[none of 1st n digits are 6]=(0.9)n. Indpt. digits?
