Find a joint distribution for y1 y2 y3 that satisfies the


The point of this exercise is to show that the sequence of PMFs for a Bernoulli counting process does not specify the process. In other words, knowing that N(t) satisfies the binomial distribution for all t does not mean that the process is Bernoulli. This helps us understand why the second definition of a Poisson pro- cess requires stationary and independent increments as well as the Poisson distribution for N(t).

(a) Let Y1, Y2, Y3, ... be a sequence of binary rv s in which each rv is 0 or 1 with equal probability. Find a joint distribution for Y1, Y2, Y3 that satisfies the binomial distribution, pN(t)(k) = t )2-t for t = 1, 2, 3 and 0 ≤ k ≤ t, but for which Y1, Y2, Y3 are not independent.

One simple solution for this contains four 3-tuples with probability 1/8 each, two 3-tuples with probability 1/4 each, and two 3-tuples with probability 0. Note that by making the subsequent arrivals IID and equiprobable, you have an example where N(t) is binomial for all t but the process is not Bernoulli. Hint: Use the binomial for t = 3 to find two 3-tuples that must have probability 1/8. Combine this with the binomial for t = 2 to find two other 3-tuples that must have probability 1/8. Finally, look at the constraints imposed by the binomial distribution on the remaining four 3-tuples.

(b) Generalize (a) to the case where Y1, Y2, Y3 satisfy Pr{Yi = 1} = p and Pr{Yi = 0} = 1 - p. Assume p <>1/2 and find a joint distribution on Y1, Y2, Y3 that satisfies the binomial distribution, but for which the 3-tuple (0, 1, 1) has zero probability.

More generally yet, view a joint PMF on binary t-tuples as a non-negative vector ina 2t dimensional vector space. Each binomial probability pN(τ )(k) = τ )pk(1 - p)τ -k constitutes a linear constraint on this vector. For each τ , show that one of these constraints may be replaced by the constraint that the components of the vector sum to 1.

(d) Using (c), show that at most (t + 1)t/2 + 1 of the binomial constraints are linearly independent. Note that this means that the linear space of vectors satisfying these binomial constraints has dimension at least 2t - (t + 1)t/2 - 1. This linear space has dimension 1 for t = 3, explaining the results in (a) and (b). It has a rapidly increasing dimension for t > 3, suggesting that the binomial constraints are relatively ineffectual for constraining the joint PMF of a joint distribution. More work is required for the case of t > 3 because of all the inequality constraints, but it turns out that this large dimensionality remains.

Text Book: Stochastic Processes: Theory for Applications By Robert G. Gallager.

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Advanced Statistics: Find a joint distribution for y1 y2 y3 that satisfies the
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