let x have a weibull distribution with the pdf below axa 1e xba x 2 0 ba x 0 fx 4 b verify that

Name: let x have a weibull distribution with the pdf below axa 1e xba x 2 0 ba x 0 fx 4 b verify that bt 1 hint in the integral for ex make the change of variable y so that x by la using the subst 50806
Uploaded: 2023-04-13T21:36:29-08:00
Duration: 4 min 59 s
Channel: Numerade
Description: let x have a weibull distribution with the pdf below axa 1e xba x 2 0 ba x 0 fx 4 b verify that bt 1 hint in the integral for ex make the change of variable y so that x by la using the subst 50806

In this information we have for given information FX and capital FX B, PDF and CDF. Thus the mea

Alright, so we have this integral that we want to analyze, which is the integral from negative i

Hello in the following question Y is equal to X by p hold to the power alpha that is x to the po

All right, so in this problem, we will be showing that the function, which is, so let, let, we w

Hello students, in this question, first of all, we have to check, we need to check if the integr

Question

Let X have a Weibull distribution with the pdf below: f(x; a, B) = (a/B)(x/B)^(a-1)e^(-(x/B)^a) for x â‰¥ 0, 0 otherwise. Verify that E(X) = BÎ“(1 + 1/a). [Hint: In the integral for E(X), make the change of variable y so that x = By^(1/a). Using the substitution, Y = (x/B)^(1/a), we have dy = (1/a)(x/B)^(1/a - 1)(1/B) dx. Now we can simplify E(X) as follows: E(X) = âˆ«[0,âˆž] x f(x; a, B) dx = âˆ«[0,âˆž] x (a/B)(x/B)^(a-1)e^(-(x/B)^a) dx = âˆ«[0,âˆž] (By^(1/a))(a/B)(By^(1/a)/B)^(a-1)e^(-(By^(1/a)/B)^a) (1/a)(By^(1/a - 1))(1/B) dy = âˆ«[0,âˆž] (a/B)(By^(1/a))^a e^(-(By^(1/a))^a) (1/a)(By^(1/a - 1))(1/B) dy = (a/B)(1/a)(1/B) âˆ«[0,âˆž] y^a e^(-y^a) y^(1/a - 1) dy = (1/B^2) âˆ«[0,âˆž] y^(a + 1/a - 1) e^(-y^a) dy = (1/B^2) âˆ«[0,âˆž] y^(1 + 1/a) e^(-y^a) dy = (1/B^2) Î“(1 + 1/a).

          Let X have a Weibull distribution with the pdf below:
f(x; a, B) = (a/B)(x/B)^(a-1)e^(-(x/B)^a) for x â‰¥ 0, 0 otherwise.

Verify that E(X) = BÎ“(1 + 1/a).

[Hint: In the integral for E(X), make the change of variable y
so that x = By^(1/a).

Using the substitution, Y = (x/B)^(1/a), we have
dy = (1/a)(x/B)^(1/a - 1)(1/B) dx.

Now we can simplify E(X) as follows:
E(X) = âˆ«[0,âˆž] x f(x; a, B) dx
     = âˆ«[0,âˆž] x (a/B)(x/B)^(a-1)e^(-(x/B)^a) dx
     = âˆ«[0,âˆž] (By^(1/a))(a/B)(By^(1/a)/B)^(a-1)e^(-(By^(1/a)/B)^a) (1/a)(By^(1/a - 1))(1/B) dy
     = âˆ«[0,âˆž] (a/B)(By^(1/a))^a e^(-(By^(1/a))^a) (1/a)(By^(1/a - 1))(1/B) dy
     = (a/B)(1/a)(1/B) âˆ«[0,âˆž] y^a e^(-y^a) y^(1/a - 1) dy
     = (1/B^2) âˆ«[0,âˆž] y^(a + 1/a - 1) e^(-y^a) dy
     = (1/B^2) âˆ«[0,âˆž] y^(1 + 1/a) e^(-y^a) dy
     = (1/B^2) Î“(1 + 1/a).

let x have a weibull distribution with the pdf below axa 1e xba x 2 0 ba x 0 fx 4 b verify that bt 1 hint in the integral for ex make the change of variable y so that x by la using the subst 50806

Added by Mark A.

Question

Please give Ace some feedback