Final answer:
The expectation of a binomially distributed variable X with parameters n and p is np. Increasing X by one yields a new variable Y whose expectation is E[Y] = E[X] + 1 = np + 1. However, the original statement claiming E[Y] equals (n+1)p / [1 - (1-p)^(n+1)] requires further proof or context, as it is not evident the expressions are equivalent.
Step-by-step explanation:
To prove that for a random variable X with a binomial distribution with parameters n and p, the expected value of (X+1) is equal to (n+1)p / [1 - (1-p)(n+1)], we start by noting the expected value E[X] for a binomially distributed variable is np. This property is derived from the definition of expected value as the sum over all possible outcomes of the product of the outcome value and its probability.
The expression (X+1) can be considered as a new variable, let's denote it Y. The expected value of Y, E[Y], is the sum of the expected value of X and 1, as each occurrence of X in the probability distribution is increased by one. Thus, E[Y] = E[X] + 1 = np + 1. However, the original claim is that E[Y] = (n+1)p / [1 - (1-p)(n+1)], which is not obviously the same as np + 1. Therefore, additional steps would be required to prove that these two expressions are actually equivalent, which may involve complex algebraic manipulations or a deeper understanding of the binomial distribution's properties.
Given the above, the proof statement seems to be incorrect unless there is additional context that redefines the meaning of the random variable Y or incorporates other factors still to be considered.
The complete question is: Show that for a random variable X with a binomial distribution with parameters n and p, the expected value of (X+1)/(1) is equal to (n+1)p / [1 - (1-p)^(n+1)]. is: