Iecuhk engg2430c tutorial 4 february 12 2013 69

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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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This test prep was uploaded on 03/31/2014 for the course ENGG 2430C at The Chinese University of Hong Kong.

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