Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Use the definitions of expectation and variance.
- Expectation
[tex]E(X) = \displaystyle \int_{-\infty}^\infty x f_X(x) \, dx = \frac14 \int_0^\infty x e^{-x/4} \, dx[/tex]
Integrate by parts,
[tex]\displaystyle \int_a^b u \, dv = uv \bigg|_a^b - \int_a^b v \, du[/tex]
with
[tex]u = x \implies du = dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]
Then
[tex]E(X) = \displaystyle \frac14 \left(\left(-4x e^{-x/4}\right)\bigg|_0^\infty + 4 \int_0^\infty e^{-x/4} \, dx\right)[/tex]
[tex]E(X) = \displaystyle \int_0^\infty e^{-x/4} \, dx = \boxed{4}[/tex]
(since the integral of the PDF is 1, and this integral is 4 times that)
- Variance
[tex]V(X) = E\bigg((X - E(X))^2\bigg) = E(X^2) - E(X)^2[/tex]
Compute the so-called second moment.
[tex]E(X^2) = \displaystyle \int_{-\infty}^\infty x^2 f_X(x)\, dx = \frac14 \int_0^\infty x^2 e^{-x/4} \, dx[/tex]
Integrate by parts, with
[tex]u = x^2 \implies du = 2x \, dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]
Then
[tex]E(X^2) = \displaystyle \frac14 \left(\left(-4x^2 e^{-x/4}\right)\bigg|_0^\infty + 8 \int_0^\infty x e^{-x/4} \, dx\right)[/tex]
[tex]E(X^2) = 8 E(X) = 32[/tex]
and the variance is
[tex]V(X) = 32 - 4^2 = \boxed{16}[/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
