Let X denote an exponential random variable with unknown parameter λ>0 . Let Y=I(X>5) , the indicator that X is larger than 5 .
Recall the definition of the indicator function here is
I(X>5)={1ifX>50ifX≤5.
We think of Y as a censored version of the Exponential random variable X : we cannot directly observe X , but we are able to gather some information about it (in this case, whether or not X is larger than 5 .)
Observe that Y is a Bernoulli random variable. Thus, the statistical model for Y can be written ({0,1},{Ber(f(λ))}λ>0) for some function f of λ . What is f(λ) ?
(Type lambda for λ . Use the help button below for help with formula input).
f(λ)=
1-e^(-λ*5)
To determine the function f(λ) in the statistical model for Y, we need to find the probability that Y takes the value 1 (indicating X > 5) given the parameter λ. In other words, we need to find P(Y = 1|λ).
Since Y is a Bernoulli random variable, we know that P(Y = 1|λ) = f(λ), where f(λ) represents the probability of success for the Bernoulli distribution.
In this case, Y takes the value 1 if X > 5, and 0 otherwise. We can express this as:
P(Y = 1|λ) = P(X > 5|λ)
Since X is an exponential random variable with parameter λ, we can use the cumulative distribution function (CDF) of the exponential distribution to find this probability.
The CDF of the exponential distribution is given by:
F(x;λ) = 1 - e^(-λx)
Using this CDF, we can calculate:
P(X > 5|λ) = 1 - P(X ≤ 5|λ) = 1 - F(5;λ) = 1 - (1 - e^(-5λ)) = e^(-5λ)
Therefore, f(λ) = e^(-5λ).
To find the function f(λ), we need to determine the probability distribution of Y, which is a Bernoulli random variable.
In this case, Y takes two possible values: 0 when X ≤ 5 and 1 when X > 5.
The probability of Y = 1, P(Y=1), can be calculated as the complement of the probability of Y = 0, P(Y=0).
Since Y is determined by whether X is larger than 5, we can express P(Y=1) as P(X > 5).
For an exponential random variable X with parameter λ, the probability that X > 5 can be calculated using the cumulative distribution function (CDF) of the exponential distribution.
The CDF of an exponential distribution is F(x) = 1 - e^(-λx), where λ is the parameter and x is the value we are evaluating the CDF at.
In this case, we want to calculate P(X > 5), so we substitute x = 5 into the CDF:
F(5) = 1 - e^(-λ(5))
Therefore, the function f(λ) can be expressed as:
f(λ) = 1 - e^(-λ(5))
So, the answer is f(λ) = 1 - e^(-λ(5)).