Show that if the random variable X has the geometric distribution with parameter p, and j is a p

Show that if the random variable X has the geometric...

Key Concepts

Geometric Series Summation

The derivation of the survival function for the geometric distribution involves summing an infinite geometric series. Recognizing that P(X ? j) is the sum of p(1-p)^(k-1) from k = j to infinity, one can factor out (1-p)^(j-1) and apply the formula for the sum of a geometric series. This simplification yields P(X ? j) = (1-p)^(j-1), thereby showcasing the utility of geometric series in solving problems related to probability distributions.

Geometric Distribution

The geometric distribution models the number of independent Bernoulli trials needed to get the first success, with each trial having a success probability p. It is a discrete distribution defined on the set of positive integers, and its defining property is that each trial is independent of the previous ones.

Probability Mass Function (PMF)

The probability mass function (PMF) of the geometric distribution is given by P(X = k) = p(1-p)^(k-1), where k is a positive integer. This function indicates the probability that the first success occurs exactly on the kth trial, with the term (1-p)^(k-1) representing the probability of having k-1 consecutive failures before the success.

Survival Function

The survival function, in the context of the geometric distribution, represents the probability that the random variable X takes on a value greater than or equal to a certain integer j. It is computed as P(X ? j), which, for the geometric distribution, can be derived by summing the probabilities of X taking each value from j to infinity.

Transcript

00:01 Okay, so for this problem, we are given that random variable x has geometric distribution with probability p, and then we're given that j is some positive integer.

00:30 It's an integer.

00:38 Okay, and then we want to prove that the probability that are geometrically distributed random variable is greater than equal to j is equal to 1 minus the probability of occurring to the power of j minus 1.

01:03 So we're actually going to determine the probability of the first j minus 1 probabilities here, right? so we're looking at the probability that x is less than j, and then we'll use the complement rule to figure out the probability of x greater than or equal to j here.

01:29 So the probability that x is less than j is going to be the sum from 1 to j minus 1 of the probability that x equals that value, right? that's just by the definition.

02:02 Okay, we can break this down to the sum from 1 to j minus 1, 1 minus p to k minus 1 times, p and this is just by the definition of a geometric probability so i'm just using the definition of a geometric probability here okay we can pull this p out and bring it to the front right so that's going to be then equal to p times the sum from 1 to j minus 1 1 to the power of k minus 1 so i'm going to move on to a clean page here.

03:29 But now i'm just going to change the sum index to sort of simplify it a little bit, right? so we still have the p out front.

03:40 But instead of k, we're going to start at i is equal to zero and run that to j minus two.

03:52 What that gives us here then is it's going to be 1 minus p just to the power.

03:58 Of i okay and so we did is we let i be equal to k minus one right now we're going to also use just the definition of a geometric sum here to get that this is equal to p times one minus p the power of j minus one minus one over one minus p minus one right so the definition of a geometric sum here states that if you have a sum from that runs from 0 to n, let's call it x to the power of i here, that is equal to x to the power of n plus 1 minus 1 or x minus 1.

05:02 So it's just this definition that i've applied at this step here...

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