Math  /  Calculus

Question3.32. Në qoftë se densiteti i XX është f(x)=1,0<x<1f(x)=1, \quad 0<x<1 gjeni E[etX]E\left[e^{t X}\right]. Kryeni llogaritjet për të gjetur E(Xn)E\left(X^{n}\right).

Studdy Solution
Evaluate the integral:
E(Xn)=01xndx E\left(X^n\right) = \int_{0}^{1} x^n \, dx
To solve this, we find the antiderivative of xn x^n :
xndx=xn+1n+1+C \int x^n \, dx = \frac{x^{n+1}}{n+1} + C
Now, evaluate from 0 to 1:
E(Xn)=[xn+1n+1]01 E\left(X^n\right) = \left[ \frac{x^{n+1}}{n+1} \right]_{0}^{1} E(Xn)=1n+1n+10n+1n+1 E\left(X^n\right) = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} E(Xn)=1n+1 E\left(X^n\right) = \frac{1}{n+1}
The expected values are:
E[etX]=1t(et1) E\left[e^{tX}\right] = \frac{1}{t} (e^{t} - 1) E(Xn)=1n+1 E\left(X^n\right) = \frac{1}{n+1}

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord