lengths of the different pieces are independent, and the length of each piece is distributed according to the same PDF fX(x). Let R be the length of the piece that includes the dot. Determine the expected value of R in each of the following cases.
In each part below, express your answer in terms of λ using standard notation. Enter 'lambda' for λ.
Suppose that fX(x)={λe−λx,0,x≥0,x<0.
E[R]=- unanswered
Suppose that fX(x)={λ3x2e−λx2,0,x≥0,x<0.
E[R]=- unanswered
To determine the expected value of R in each case, we need to calculate the integral of R multiplied by the probability density function fX(x) with respect to x, and evaluate it from 0 to infinity.
1. In the first case, where fX(x) = λe^(-λx) when x ≥ 0 and x < 0 otherwise:
E[R] = ∫(0 to ∞) R * λe^(-λx) dx
To solve this integral, we use integration by parts. Let u = R and dv = λe^(-λx) dx. Then du = dR and v = -e^(-λx) / λ.
E[R] = [-R * e^(-λx) / λ] from 0 to ∞ + ∫(0 to ∞) e^(-λx) / λ dx
Since e^(-λx) approaches 0 as x approaches infinity, the first term on the right side of the equation is 0. Therefore, we are left with:
E[R] = ∫(0 to ∞) e^(-λx) / λ dx
To solve this integral, we can make use of the property of the exponential distribution:
∫(0 to ∞) e^(-μx) dx = 1 / μ
By replacing λ with μ, we can rewrite the equation as:
E[R] = 1 / λ
Therefore, the expected value of R in the first case is 1 / λ.
2. In the second case, where fX(x) = λ3x^2e^(-λx^2) when x ≥ 0 and x < 0 otherwise:
E[R] = ∫(0 to ∞) R * λ3x^2e^(-λx^2) dx
Since R is a constant, we can move it outside the integral:
E[R] = λ3 * ∫(0 to ∞) x^2e^(-λx^2) dx
This integral does not have a closed-form solution. However, it can be evaluated numerically using methods such as numerical integration or approximations.
Therefore, the expected value of R in the second case is λ3 * (∫(0 to ∞) x^2e^(-λx^2) dx).
To determine the expected value of R in each case, we need to calculate the integral of the product of the length of the piece and its probability density function (PDF) over the entire range.
1. Case 1: fX(x) = λe^(-λx), 0 < x.
The PDF fX(x) represents an exponential distribution with rate parameter λ.
To find the expected value of R, we integrate the product of x (length of the piece) and fX(x) over the range from 0 to infinity:
E[R] = ∫(0 to ∞) x * fX(x) dx
Substituting the given PDF, we have:
E[R] = ∫(0 to ∞) x * (λe^(-λx)) dx.
To solve this integral, we can use integration by parts. Let u = x and dv = λe^(-λx) dx.
Differentiating u with respect to x, we get du = dx, and integrating dv, we have v = -e^(-λx).
Using the integration by parts formula, ∫u dv = uv - ∫v du, we can rewrite the integral as:
E[R] = [-xe^(-λx)] (0 to ∞) + ∫(0 to ∞) e^(-λx) dx.
Evaluating the limits at 0 and infinity, the first term becomes 0. We are then left with:
E[R] = ∫(0 to ∞) e^(-λx) dx.
This integral represents the area under the curve of the exponential distribution, which is equal to 1. Therefore, we can conclude that:
E[R] = 1 / λ.
2. Case 2: fX(x) = λ^3x^2e^(-λx^2), 0 < x.
The PDF fX(x) represents a chi-squared distribution with 2 degrees of freedom and scale parameter λ.
To find the expected value of R, we integrate the product of x (length of the piece) and fX(x) over the range from 0 to infinity:
E[R] = ∫(0 to ∞) x * fX(x) dx
Substituting the given PDF, we have:
E[R] = ∫(0 to ∞) x * (λ^3x^2e^(-λx^2)) dx.
We can simplify this integral by performing a change of variable. Let u = λx^2, then du = 2λxdx.
Therefore, the integral becomes:
E[R] = (1 / 2λ) * ∫(0 to ∞) u * e^(-u) du.
The integral ∫u * e^(-u) du is the Gamma function, denoted as Γ(2). The Gamma function is equal to 1, so we have:
E[R] = (1 / 2λ) * Γ(2),
where Γ(2) is equal to 1.
Simplifying further, we find:
E[R] = 1 / (2λ).
To summarize:
- For Case 1, where fX(x) = λe^(-λx), the expected value of R is E[R] = 1 / λ.
- For Case 2, where fX(x) = λ^3x^2e^(-λx^2), the expected value of R is E[R] = 1 / (2λ).