\displaystyle f(x) = \frac \lambda 2 e^{-\lambda|x-\mu|} = \cases { \frac \lambda 2 e^{\lambda (x-\mu)} &\text{dla } x < \mu\cr \frac \lambda 2 e^{-\lambda (x-\mu)} &\text{dla } x \geq \mu\cr }
\displaystyle E(X) = \int_{-\infty}^{\mu} {\frac \lambda 2 xe^{\lambda (x-\mu)}} + \int_{\mu}^{\infty} {\frac \lambda 2 x e^{-\lambda (x-\mu)}} = \text{dalej mi się nie chce narazie :P}
E(X) nie istnieje, bo całka \displaystyle \int_{-\infty}^{\infty}{ \frac 1 {\pi} \frac \lambda { \lambda^2 + (x-\mu)^2 } } = \frac {\lambda} {\pi} \bigg (\frac 1 2 log( (x-\mu)^2+\lambda^2) + \frac 1 {\lambda} \mu \tan^{-1}(\frac{x-\mu}{\lambda}) \bigg ) jest rozbieżna.
f(x) = xe^{-x}
\displaystyle E(X) = \int_0^{\infty} {x^2e^{-x}} = -x^2e^{-x}\bigg|_0^{\infty} + 2\int_0^{\infty} {xe^{-x}} = -x^2e^{-x}\bigg|_0^{\infty} - 2xe^{-x}\bigg|_0^{\infty} + 2e^{-x}\bigg|_0^{\infty} = 2
\displaystyle E(X^2) = \int_0^{\infty} {x^3e^{-x}} = -x^3e^{-x}\bigg|_0^{\infty} -3x^2e^{-x}\bigg|_0^{\infty} - 6xe^{-x}\bigg|_0^{\infty} + 6e^{-x}\bigg|_0^{\infty} = 6
D^2(X) = E(X^2) - E(X)^2 = 6 - 4 = 2
f(x) =\cases { x&\text{dla } 0\leq x \leq1\cr 2-x&\text{dla } 1\leq x\leq 2\cr 0&\text{otherwise}\cr }
\displaystyle E(X) = \int_{-\infty}^{\infty} {x \cdot f(x)} = \int_0^1 {x^2} + \int_1^2{(2x - x^2)} = \frac {x^3} 3 |_0^1 + x^2 |_1^2 - \frac {x^3} 3 |_1^2 = \frac 1 3 + 4 - 1 - \frac 8 3 + \frac 1 3 = 1
\displaystyle E(X^2) = \int_{-\infty}^{\infty} {x^2 \cdot f(x)} = \int_0^1 {x^3} + \int_1^2{(2x^2 - x^3)} = \frac {x^4} 4 |_0^1 + \frac {2x^3} 3 |_1^2 - \frac {x^4} 4 |_1^2 = \frac 1 4 + \frac {16} 3 - \frac 2 3 - \frac {16} 4 + \frac 1 4 = \frac 7 6
\displaystyle D^2(X) = E(X^2) - E(X)^2 = \frac 7 6 - 1 = \frac 1 6
F(x) =\cases { 0&\text{dla } x\leq 1\cr 2-\frac 2 x&\text{dla } 1<x\leq a\cr 1&\text{dla } x>a\cr }
f(x) = F'(x) = \cases{ 0 & \text{dla } x\leq 1\cr \frac 2 {x^2} & \text{dla } 1<x\leq a\cr 0 & \text{dla } x>a\cr }
Notabene a musi być równe 2, bo w przeciwnym wypadku f(x) nie sumuje się do jedynki.
E(X) = \int {xf(x) dx} = \int_1^a {\frac 2 x} dx = 2 log(a)
E(X^2) = \int {x^2f(x) dx} = \int_1^a 2 dx = 2x |_1^a = 2a - 2
D^2(X) = E(X^2) - E(X)^2 = 2a-2 -4log^2(a)
f(x) = 3x^2
\displaystyle E(X) = \int_0^1{3x^3} = \frac 3 4 x^4 |_0^1 = \frac 3 4
\displaystyle E(X^2) = \int_0^1{3x^4} = \frac 3 5 x^5 |_0^1 = \frac 3 5
\displaystyle D^2(X) = E(X^2) - E(X)^2 = \frac 3 5 - \frac 9 {16} = \frac 3 {80}