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

STEP 1

1. The random variable X X has a uniform distribution on the interval (0,1) (0, 1) .
2. The probability density function (pdf) of X X is f(x)=1 f(x) = 1 for 0<x<1 0 < x < 1 .
3. We need to find the expected value E[etX] E\left[e^{tX}\right] .
4. We also need to find the expected value E(Xn) E\left(X^n\right) .

STEP 2

1. Set up the integral for E[etX] E\left[e^{tX}\right] .
2. Evaluate the integral for E[etX] E\left[e^{tX}\right] .
3. Set up the integral for E(Xn) E\left(X^n\right) .
4. Evaluate the integral for E(Xn) E\left(X^n\right) .

STEP 3

Set up the integral for E[etX] E\left[e^{tX}\right] :
E[etX]=01etxf(x)dx E\left[e^{tX}\right] = \int_{0}^{1} e^{tx} \cdot f(x) \, dx E[etX]=01etx1dx E\left[e^{tX}\right] = \int_{0}^{1} e^{tx} \cdot 1 \, dx

STEP 4

Evaluate the integral:
E[etX]=01etxdx E\left[e^{tX}\right] = \int_{0}^{1} e^{tx} \, dx
To solve this, we find the antiderivative of etx e^{tx} :
etxdx=1tetx+C \int e^{tx} \, dx = \frac{1}{t} e^{tx} + C
Now, evaluate from 0 to 1:
E[etX]=[1tetx]01 E\left[e^{tX}\right] = \left[ \frac{1}{t} e^{tx} \right]_{0}^{1} E[etX]=1tet1te0 E\left[e^{tX}\right] = \frac{1}{t} e^{t} - \frac{1}{t} e^{0} E[etX]=1t(et1) E\left[e^{tX}\right] = \frac{1}{t} (e^{t} - 1)

STEP 5

Set up the integral for E(Xn) E\left(X^n\right) :
E(Xn)=01xnf(x)dx E\left(X^n\right) = \int_{0}^{1} x^n \cdot f(x) \, dx E(Xn)=01xn1dx E\left(X^n\right) = \int_{0}^{1} x^n \cdot 1 \, dx

STEP 6

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}

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