Problem 10 The memoryless property Suppose F has geometric...

Problem 10 The memoryless property Suppose F has geometric distribution on {0, 1, 2, dots} as in Problem 9. a) Show that for every $k ge 0$ $P(F - k = m mid F ge k) = P(F = m)$, $m = 0, 1, dots$ b) What is the corresponding characterization of the geometric ($p$) distribution on {1, 2, dots}?

Transcript

00:01 In the asked question, we suppose f is a random variable that had geometric distribution follows a geometric distribution on the interval from 0 to infinity and it can't take any integer value because the geometric distribution is a distribution of a discrete random variable f is a discrete random variable for party a is required to show that for every k is greater than equal 0 the probability for f minus k to be equal m given that f is already greater than equals k equals the probability of f equals m for m is any injured number starting from 0 we can prove that by getting the left hand side starting from the left hand side we can go to prove that it equals right hand side the left hand side is a conditional probability by definition it equals the intersection between the probability of the intersection between these two events which is f minus k equals m and if is greater than equals k divided by the probability of the conditioning event if to be greater than equals k which equals the probability for f to be equal k plus m we can get k in the other side with a negative sign then it's k plus m and f to be greater than equals k divided by the denominator let's focus on the denominator now what is the probability of f to be equal to k plus any integer value like this given that it's already greater than equals k we can notice that here we added a value of k then this is of course a value greater than k then the left event this event contains already this event or satisfies already this event because m k plus m is already greater than k then this is just the bmf which is the probability for f equals k plus m because this event satisfies this event we can take only the left event divided by the probability of f to be greater than equals k we can rewrite it as one minus the probability of f to be smaller than k using the complement event and what is the probability of f to be smaller than k we can add equal sign but account for this addition then it's one minus the probability for f to be smaller than or equals k you can get it here in brackets smaller than or equals k but i have added the probability for f equals k i should remove it now this is equivalent to this why i have done that because this is the cdf of the random variable f add value now we just substitute here what is the bdf sorry the bmf what is the bmf add value k plus m we substitute in the equation the bmf of the geometric distribution starting from 0 it's 1 minus the probability 1 minus b or the parameter to the power of k plus m multiplied by b divided by 1 minus the cdf at value equals k what is the cdf it's 1 minus 1 minus b to the power of k plus 1 minus minus what is the bmf at value equal k it's 1 minus p to the bar of k multiplied by b this equals 1 minus 1 gives 0 then minus minus gives positive and minus minus minus gives positive it's 1 minus b to the bar of k plus m multiplied by b divided by 1 minus b to the bar of k plus 1 plus 1 minus b to the bar of k plus 1 plus 1 minus b to the bar of k plus multiply it by b you can notice that from that denominator we can get 1 minus p to the bar of k as a common factor then it's 1 minus b to the bar of k multiply it by 1 minus b to the bar of k multiplied by 1 minus b to the bar of b divided by 1 minus b to the bar of k applied by 1 minus b to the bar of 1 which is 1 minus b plus b equals we can remove 1 minus p to sub power of k with this and here minus p…

Question

Problem 10 The memoryless property Suppose F has geometric distribution on {0, 1, 2, dots} as in Problem 9. a) Show that for every $k ge 0$ $P(F - k = m mid F ge k) = P(F = m)$, $m = 0, 1, dots$ b) What is the corresponding characterization of the geometric ($p$) distribution on {1, 2, dots}?

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