The lifetime (in thousands of hours) X of a light bulb has pdf g(x)= \lambda e^{-\lambda x}, \hspace{3mm} x\geq 0 for some unknown \lambda >0. Find \mathbb E[X].
\mathbb E[X]=\quad
To find the expected value of X, we can use the formula:
E[X] = ∫ x * g(x) dx
Plugging in the given pdf g(x) = λe^(-λx), we have:
E[X] = ∫ x * λe^(-λx) dx
To solve this integral, we can use integration by parts. Let u = x and dv = λe^(-λx) dx. Then, du = dx and v = -e^(-λx)/λ.
Using the formula for integration by parts:
∫ u * dv = uv - ∫ v * du
We have:
E[X] = [-x * e^(-λx)/λ] - ∫ (-e^(-λx)/λ) dx
= [-x * e^(-λx)/λ] + [e^(-λx)/λ^2] + C
where C is the constant of integration.
To find the limits of integration, recall that the pdf g(x) is defined for x ≥ 0. Therefore, the integral is evaluated from 0 to ∞.
E[X] = [0 * e^(-λ*0)/λ] + [e^(-λ*0)/λ^2] - [-∞ * e^(-λ*∞)/λ] - [e^(-λ*∞)/λ^2]
Since e^(-λ∞) approaches 0 as λ approaches infinity, the last two terms in the equation above simplify to 0.
E[X] = 0 + (1/λ^2) - 0 - 0
= 1/λ^2
Therefore, the expected value of X is 1/λ^2.
To find the expected value of X, denoted as E[X], we can use the following formula:
E[X] = ∫x*g(x)dx
Where g(x) is the probability density function (pdf) of X.
In this case, the pdf g(x) is given as g(x) = λe^(-λx), x ≥ 0.
Substituting the pdf into the formula, we have:
E[X] = ∫x * λe^(-λx) dx
To solve this integral, we can use integration by parts.
Let u = x and dv = λe^(-λx) dx
Then, du = dx and v = -e^(-λx)/λ
Using the integration by parts formula: ∫u dv = uv - ∫v du, we can evaluate the integral.
E[X] = [-xe^(-λx)/λ] - ∫(-e^(-λx)/λ) dx
E[X] = [-xe^(-λx)/λ] + [∫e^(-λx)/λ dx]
The integral on the right-hand side can be evaluated as follows:
∫e^(-λx)/λ dx = -[e^(-λx)/λ^2]
Substituting this result back into our equation, we have:
E[X] = [-xe^(-λx)/λ] - [-e^(-λx)/λ^2] + C
E[X] = (-xe^(-λx) + e^(-λx)/λ^2) + C
Since we want to find the expected value of X, we need to evaluate this expression at the limits of integration (from 0 to infinity):
E[X] = lim(x → ∞) of (-xe^(-λx) + e^(-λx)/λ^2) - (-0e^(-λ(0)) + e^(-λ(0))/λ^2)
E[X] = lim(x → ∞) of (-xe^(-λx) + e^(-λx)/λ^2) - 0 + 1/λ^2
E[X] = lim(x → ∞) of -xe^(-λx) + e^(-λx)/λ^2 + 1/λ^2
Taking the limit as x approaches infinity, the exponential term e^(-λx) approaches 0, since λ > 0. Therefore, we have:
E[X] = 0 + 0 + 1/λ^2
E[X] = 1/λ^2
Therefore, the expected value of X is 1/λ^2.