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Thread: Continous Random Variable

  1. #1
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    Continous Random Variable

    Question : A continuous random variable has a probability density function

    f(x) = $\displaystyle \begin{cases} \frac{1}{b-a} & a<x<b \\ 0 & otherwise \end{cases}$

    Find the cumulative distribution
    find the E[X] and Var X

    Solution

    The random variable is a uniform distribution random variable

    Cumulative Distribution function
    $\displaystyle F_x (x)$ = $\displaystyle \begin{cases}0 & x<a \\ \frac{x-a}{b-a} & a \le x<b \\1 & x \ge b \end{cases}$

    But i dont know how to prove the above ......

    My work :

    $\displaystyle F_x(x)$ = $\displaystyle \int_{ - \infty}^{a} f(t) \ dt$

    =
    $\displaystyle \frac{1}{b-a} . 0$ = 0 ...............Is this correct ?

    E[X] =
    $\displaystyle \int_{- \infty}^{\infty} x . \ f(x) dx$

    = $\displaystyle \frac{1}{b-a} \int_{- \infty}^{ \infty} x \ $dx .............What should i do now integrate with $\displaystyle - \infty$ and $\displaystyle \infty$ or what


    And what about the Var X as well??
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  2. #2
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    Smile

    Quote Originally Posted by zorro View Post
    Question : A continuous random variable has a probability density function

    f(x) = $\displaystyle \begin{cases} \frac{1}{b-a} & a<x<b \\ 0 & otherwise \end{cases}$

    Find the cumulative distribution
    find the E[X] and Var X

    Solution

    The random variable is a uniform distribution random variable

    Cumulative Distribution function
    $\displaystyle F_x (x)$ = $\displaystyle \begin{cases}0 & x<a \\ \frac{x-a}{b-a} & a \le x<b \\1 & x \ge b \end{cases}$

    But i dont know how to prove the above ......

    My work :

    $\displaystyle F_x(x)$ = $\displaystyle \int_{ - \infty}^{a} f(t) \ dt$

    =
    $\displaystyle \frac{1}{b-a} . 0$ = 0 ...............Is this correct ?

    E[X] =
    $\displaystyle \int_{- \infty}^{\infty} x . \ f(x) dx$

    = $\displaystyle \frac{1}{b-a} \int_{- \infty}^{ \infty} x \ $dx .............What should i do now integrate with $\displaystyle - \infty$ and $\displaystyle \infty$ or what


    And what about the Var X as well??
    for$\displaystyle t < a$
    $\displaystyle F_x(x)$ = $\displaystyle \int_{ - \infty}^{x} f(t) \ dt = 0$ ( $\displaystyle f(t) = 0 $ for$\displaystyle t < a$ )

    for $\displaystyle a \leq t < b$
    $\displaystyle F_x(x)$ = $\displaystyle \int_{ - \infty}^{x} f(t) \ dt = \int_{ - \infty}^{a} f(t) \ dt + \int_{a}^{x} f(t) \ dt = \int_{a}^{x} \frac{1}{b-a} \ dt = \frac{x - a}{b - a}$

    for $\displaystyle t \geq b$
    $\displaystyle F_x(x)$ = $\displaystyle \int_{ - \infty}^{x} f(t) \ dt = \int_{ - \infty}^{a} f(t) \ dt + \int_{a}^{b} f(t) \ dt + \int_{ b}^{x} f(t) \ dt = 1$

    $\displaystyle E[X] = \int_{- \infty}^{\infty} x \cdot \ f(x) ~dx = \int_{a}^{b} x \cdot \ f(x) ~dx
    $ ($\displaystyle f(x) = 0 $ outside the interval$\displaystyle (a,b)$)

    $\displaystyle Var(X) = E[X^2] - {E[X]}^2$
    calculate $\displaystyle E[X^2] = \int_{- \infty}^{\infty} x^2 \cdot \ f(x) ~dx$
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by zorro View Post
    Question : A continuous random variable has a probability density function

    f(x) = $\displaystyle \begin{cases} \frac{1}{b-a} & a<x<b \\ 0 & otherwise \end{cases}$

    Find the cumulative distribution
    find the E[X] and Var X

    Solution

    The random variable is a uniform distribution random variable

    Cumulative Distribution function
    $\displaystyle F_x (x)$ = $\displaystyle \begin{cases}0 & x<a \\ \frac{x-a}{b-a} & a \le x<b \\1 & x \ge b \end{cases}$

    But i dont know how to prove the above ......

    My work :

    $\displaystyle F_x(x)$ = $\displaystyle \int_{ - \infty}^{a} f(t) \ dt$

    =
    $\displaystyle \frac{1}{b-a} . 0$ = 0 ...............Is this correct ?

    E[X] =
    $\displaystyle \int_{- \infty}^{\infty} x . \ f(x) dx$

    = $\displaystyle \frac{1}{b-a} \int_{- \infty}^{ \infty} x \ $dx .............What should i do now integrate with $\displaystyle - \infty$ and $\displaystyle \infty$ or what


    And what about the Var X as well??
    $\displaystyle F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi$

    Now if $\displaystyle x<a,\ f(x)=0$, so:

    $\displaystyle F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^x 0\; d\xi=0$

    Now if $\displaystyle a \le x<b,\ f(x)=1/(b-a)$, so:

    $\displaystyle F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^a 0\; d\xi+\int_{a}^x \frac{1}{b-a} \; d\xi=\int_{a}^x \frac{1}{b-a} \; d\xi$

    and because $\displaystyle f(x)=0$ for $\displaystyle x\ge b$ and is a pdf, for $\displaystyle x\ge b$

    $\displaystyle F(x)=F(b)=1$

    CB
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  4. #4
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    Thank u dedust and Captain black

    Quote Originally Posted by CaptainBlack View Post
    $\displaystyle F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi$

    Now if $\displaystyle x<a,\ f(x)=0$, so:

    $\displaystyle F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^x 0\; d\xi=0$

    Now if $\displaystyle a \le x<b,\ f(x)=1/(b-a)$, so:

    $\displaystyle F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^a 0\; d\xi+\int_{a}^x \frac{1}{b-a} \; d\xi=\int_{a}^x \frac{1}{b-a} \; d\xi$

    and because $\displaystyle f(x)=0$ for $\displaystyle x\ge b$ and is a pdf, for $\displaystyle x\ge b$

    $\displaystyle F(x)=F(b)=1$

    CB


    Thank u dedust and captain black for helping be solve this problem
    Zorro is deeply great full......... amigo's
    cheers mite
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