Warmup Problems (18.440, 2/17/2015)
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1. Use induction to prove the formula 1 + 2 + ... + n = n(n+1)
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Remember: to prove an induction argument, you need to prove (1) that the
statement is true for n + 1 whenever it is true for n AND (2) that the statement
is true for some base case, say at n = 1.
Why is the base case important? Here is a counterexample "proof" to explain
why we need the base case:
"If 1 × 2... × n = 0, then 1 × . . . × n × (n + 1) = 0 × (n + 1) = 0. Therefore,
we can conclude that n! = 0 for all n."
This proof is obviously wrong. The mistake in the proof is that we forgot to
include a base case such as "1 × 2 = 0" (this "base case" would clearly be false).
Some things to watch out for
1. Pairwise disjoint vs. other forms of "disjoint" (pairwise is strongest). e.g.,
we may have ABC = ∅, but none of AB, BC, and AC are empty:
2. Some useful set operations that can be defined in terms of union, intersection, and complementation:
• Subtraction of sets: A \ B := A(B c ) "everything that is in A but NOT in
B"
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• Symmetric difference of sets: A4B := (A ∪ B) \ AB = AB c ∪ Ac B "everything in A or in B but NOT in both"
3. Some useful facts about sets
• P (A ∪ B) = P (A) + P (B) − P (AB)
We can use a Venn diagram to prove this: basically, the P (AB) is doublecounted if we do P (A) + P (B), so we must subtract P (AB) once from this
sum to get P (A ∪ B):
• P (A ∪ B) = P (A) + P (B) if and only if A and B are "almost everywhere"
disjoint (i.e., P(AB) = 0).
• P (A \ B) = P (A) − P (B) if and only if B is "almost everywhere" a subset
of A (i.e. the part of B not contained in A has probability 0: P (A \ B) = 0
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• P (AB) = P (A)P (B) means the events are independent, by definition.
Warning: P (AB) is NOT equal to P (A)P (B) in general!
• The complement of a set depends on the sample space "S": Ac = S \ A.
• P (Ac ) = 1 − P (A). This is true regardless of S, since P (S) = 1 always.
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