Respuesta :
Answer:
Step-by-step explanation:
To find the cumulative distribution function (CDF) and probability density function (PDF) of Y, we can use the transformation method. Let's start by finding the CDF.
The CDF is defined as the probability that Y is less than or equal to a given value y. Mathematically, it can be represented as:
CDF(y) = P(Y ≤ y)
To find the CDF of Y, we need to find the distribution of Y first. Given that X has an exponential distribution with mean μ, we can express X as:
X = -ln(U) / λ
where U is a random variable uniformly distributed between 0 and 1, and λ is the rate parameter of the exponential distribution (λ = 1/μ).
Substituting this expression for X into the equation for Y:
Y = 5X⁰.⁷ = 5(-ln(U) / λ)⁰.⁷
Now, we can find the CDF of Y by evaluating the probability that Y is less than or equal to y:
CDF(y) = P(Y ≤ y) = P(5(-ln(U) / λ)⁰.⁷ ≤ y)
Let's simplify this expression:
CDF(y) = P((-ln(U) / λ)⁰.⁷ ≤ y / 5)
Since (-ln(U) / λ)⁰.⁷ is a decreasing function of U, we can rewrite the inequality as:
CDF(y) = P(U ≥ exp(-(λy/5)^(1/0.7)))
Since U is uniformly distributed between 0 and 1, P(U ≥ u) = 1 - u, where u is a given value between 0 and 1. Therefore, we can rewrite the CDF as:
CDF(y) = 1 - exp(-(λy/5)^(1/0.7))
Now, let's find the PDF of Y by taking the derivative of the CDF:
PDF(y) = d/dy [1 - exp(-(λy/5)^(1/0.7))]
PDF(y) = λ * (1/0.7) * (1/5) * (λy/5)^(1/0.7 - 1) * exp(-(λy/5)^(1/0.7))
Simplifying further:
PDF(y) = (λ/0.7) * (λ/5) * (λy/5)^(-0.3) * exp(-(λy/5)^(1/0.7))
Therefore, the CDF of Y is 1 - exp(-(λy/5)^(1/0.7)), and the PDF of Y is (λ/0.7) * (λ/5) * (λy/5)^(-0.3) * exp(-(λy/5)^(1/0.7)).
Please note that λ = 1/μ, where μ is the mean of X.
