Ejercicio 1 de procesos estocásticos

A fin de aclarar aún más los conceptos emitidos hasta el momento, veamos los siguientes ejemplos:

Ejemplo 1: Sea $https://latex.codecogs.com/svg.image?\inline x(t)=at+Y$ , siendo Y una variable aleatoria uniforme en [0,1] y a una constante. Determine:

a. La media

b. La correlación o autocorrelación

c. La covarianza o autocovarianza

d. La Potencia o valor medio cuadrático

e. La varianza

f. Coeficiente de autocorrelación

Solución:

a. La media:

$https://latex.codecogs.com/svg.image?\inline u_{x}(t)=E[x(t)]=E[at+Y]+E[Y]$

$https://latex.codecogs.com/svg.image?\inline =at+E[Y]$

Donde:

$https://latex.codecogs.com/svg.image?\inline E[Y]=\int_{0}^{1}YdY=[\frac{Y^{2}}{2}]_{0}^{1}=1/2 - 0=1/2$

En consecuencia:

$https://latex.codecogs.com/svg.image?\inline u_{x}=E[X(t)]=at+E(y)=at+1/2$

b. La correlación o autocorrelación:

$https://latex.codecogs.com/svg.image?\inline R_{x}(t_{1},t_{2})=E[x(t_{1})x(t_{2})]$

Donde:

$https://latex.codecogs.com/svg.image?\inline x(t_{1})=at_{1}+y$

$https://latex.codecogs.com/svg.image?\inline x(t_{2})=at_{2}+y$

Lo cual significa que:

$https://latex.codecogs.com/svg.image?\inline x(t_{1})x(t_{2})=[at_{1}+y][at_{2}+y]$

$https://latex.codecogs.com/svg.image?\inline =a^{2}t_{1}t_{2}+a(t_{1}+t_{2})+y+y^{2}$

Por tanto:

$https://latex.codecogs.com/svg.image?\inline R_{x}(t_{1},t_{2})=E[x(t_{1})x(t_{2})]=E[a^{2}t_{1}t_{2}+a(t_{1}+t_{2})y+y^{2}]$

$https://latex.codecogs.com/svg.image?\inline =E[a^{2}t_{1}t_{2}]+E[a(t_{1}+t_{2})y]+E[y^{2}]$

$https://latex.codecogs.com/svg.image?\inline =a^{2}t_{1}t_{2}+a(t_{1}+t_{2})E[y]+E[y^{2}]$

Donde:

$https://latex.codecogs.com/svg.image?\inline E(y)=\int_{0}^{1}ydy=[\frac{y^{2}}{2}]_{0}^{1}=1/2$

$https://latex.codecogs.com/svg.image?\inline E(y^{2})=\int_{0}^{1}y^{2}dy=[\frac{y^{3}}{3}]_{0}^{1}=1/3$

En consecuencia:

$https://latex.codecogs.com/svg.image?\inline R_{x}(t_{1},t_{2})=E[x(t_{1})x(t_{2})]=a^{2}t_{1}t_{2}+\frac{1}{2}a(t_{1}+t_{2})+\frac{1}{3}$

c. La covarianza o autocovarianza

$https://latex.codecogs.com/svg.image?\inline c_{x}(t_{1},t_{2})=E[(x(t_{1})-u_{x}(t_{1}))(x(t_{2})-u_{x}(t_{2}))]$

$https://latex.codecogs.com/svg.image?\inline =R_{x}(t_{1},t_{2})-u_{x}(t_{1})u_{x}(t_{2})$

Donde:

$https://latex.codecogs.com/svg.image?\inline R_{x}(t_{1},t_{2})=a^{2}t_{1}t_{2}+\frac{1}{2}a(t_{1}+t_{2})+\frac{1}{3}$ calculado en b.)

$https://latex.codecogs.com/svg.image?\inline u_{x}(t_{1})u_{x}(t_{2})=E[x(t_{1})]E[x(t_{2})]$

$https://latex.codecogs.com/svg.image?\inline E[at_{1}+y]E[at_{2}+y]$

$https://latex.codecogs.com/svg.image?\inline =[E[at_{1}]+E(y)][E[at_{2}]+E(y)]$

$https://latex.codecogs.com/svg.image?\inline =[at_{1}+\frac{1}{2}][at_{2}+\frac{1}{2}]$

$https://latex.codecogs.com/svg.image?\inline =a^{2}t_{1}t_{2}+\frac{1}{2}a(t_{1}+t_{2})+\frac{1}{4}$

En consecuencia

$https://latex.codecogs.com/svg.image?\inline c_{x}(t_{1},t_{2})=(a^{2}t_{1}t_{2}+\frac{1}{2}a(t_{1}+t_{2})+\frac{1}{3}-(a^{2})t_{1}t_{2}+\frac{1}{2}a(t_{1}+t_{2})+\frac{1}{4})$

$https://latex.codecogs.com/svg.image?\inline =\frac{1}{3}-\frac{1}{4}=\frac{4-3}{12}=\frac{1}{12}$

d. La potencia o valor medio cuadrático

$https://latex.codecogs.com/svg.image?\inline E[x^{2}(t)]=\int_{-\infty }^{\infty }x^{2}fx(xt)dx=R_{x}(t_{1},t_{2})$

$https://latex.codecogs.com/svg.image?\inline E[x^{2}(t)]=E[(at+y)^{2}]=E[a^{2}t^{2}+2aty+y^{2}]$

$https://latex.codecogs.com/svg.image?\inline =a^{2}t^{2}+at+1/3$

Si previamente se ha determinado $https://latex.codecogs.com/svg.image?\inline R_{x}(t_{1},t_{2})$ también se puede calcular usando la formula $https://latex.codecogs.com/svg.image?\inline R_{x}(t,t)$ , donde t1= t2 = t

e. La varianza:

DKESCORPIO

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Ejercicio 1 de procesos estocásticos

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