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**Unformatted text preview: **K) ENGG2430C tutorial 4 February 12, 2013 6/9 Example 3: Expectation and Variance
Solution:
The scalar a must satisfy
1 = ∑ pX (x ) =
x 13 2
∑x.
a x =−3 So a = 28.
We also have E[X ] = 0 because PMF is symmetric around 0.
If z ∈ {1, 4, 9}, then
√
√
pZ (z ) = pX (− z ) + pX ( z ) = z /14;
Otherwise, pZ (z ) = 0.
The variance of X is
var(X ) = E[Z ] = ∑ zpZ (z ) =
z (IE@CUHK) ENGG2430C tutorial 4 z2
∑ 14 = 7.
z ∈{1,4,9}
February 12, 2013 7/9 Puzzle
Example
A jar begins with one bacteria. Every minute, every bacteria turns into 0,
1, 2, or 3 bacteria with a probability of 25% for each case (dies, does
nothing, splits into 2, or splits into 3). What is the probability that the
bacteria population eventually dies out? (IE@CUHK) ENGG2430C tutorial 4 February 12, 2013 8/9 Puzzle
Solution:
If p is the probability that a single bacteria’s descendants will die out
eventually, the probability that n bacteria’ descendants will all die out
eventually must be p n , since each bacteria is independent of every
other bacteria. In other words,
P(eventually die out | n bacteria) = p n .
Also, the probability that a single bacteria’s descendants will die out
must be independent of time. Conditioning on the number of bacteria
at the end of the 1st min:
3 p =
= ∑ P(eventually die out | i bacteria) · P(i bacteria after the 1st min) i =0
0 (p + p 1 + p 2 + p 3 ) · 0.25, which gives us p =
(IE@CUHK) √
2 − 1. ENGG2430C tutorial 4 February 12, 2013 9/9...

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