Let the sequence Xn, n = 1, 2, 3, ..., be a Bernoulli process with parameter P(Xn = 1) = p for all n >= 1. Let be the time when a value of 0 is first observed: U = min{n: Xn = 0} Then, the random variable is:
A) Geometric with parameter
B) Geometric with parameter
C) None of the above
Geometric with parameter 1-p
It would help if you would proofread your work before you post it.
A and B are identical.
jiskha.com/questions/1891168/let-the-sequence-xn-n-1-2-3-be-a-bernoulli-process-with-parameter-p-xn-1-p
To determine the random variable of the sequence U, we can analyze the problem step by step.
First, let's understand what the sequence Xn represents. The Bernoulli process describes a sequence of independent binary events, where each event can have one of two outcomes: success (denoted by 1) or failure (denoted by 0). The parameter p represents the probability of a success (Xn = 1) occurring.
Now, let's define the random variable U as the time when a value of 0 is first observed in the sequence Xn. This means that U represents the number of trials needed until the first failure occurs (Xn = 0). In other words, U is the number of successes before the first failure.
To determine the distribution of U, we can consider the probability of getting a specific number of successes before the first failure.
If U = k, it means that we had k-1 successes followed by a failure. The probability of this specific sequence happening is given by: P(X1 = 1, X2 = 1, ..., Xk-1 = 1, Xk = 0) = (p^(k-1))(1-p).
Therefore, the probability distribution of U follows a geometric distribution with parameter (1-p). In a geometric distribution, the random variable represents the number of trials needed to achieve the first success, and the parameter represents the probability of success on any given trial.
In this case, U represents the number of successes before the first failure, and the parameter (1-p) represents the probability of failure on any given trial.
Hence, the correct answer is A) Geometric with parameter (1-p).