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Thread: Continuous Random Variable Proof

  1. #1
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    Continuous Random Variable Proof

    Question: If X is a random variable for which $\displaystyle P(X \geq 0)=0$ and $\displaystyle |\mu_x|<\infty$, show that $\displaystyle P(X<\mu t) \geq 1 - \frac{1}{t}$ for every $\displaystyle t \geq 1$.

    My TA told me to use the following information:
    $\displaystyle \mu=\int_{-\infty}^{\infty}(x(f(x))dx$
    Since $\displaystyle P(X \geq 0)=0$:
    $\displaystyle \mu=\int_{0}^{\infty}(x(f(x))dx$
    $\displaystyle \mu=\int_{0}^{|\mu t|}(x(f(x))dx + \int_{|\mu t|}^{\infty}(x(f(x))dx$

    My own work:
    We know that:
    $\displaystyle P(X<\mu t)=\int^{\mu t}_{-\infty}(f(x))dx$
    Since $\displaystyle P(X \geq 0)=0$:
    $\displaystyle P(X<|\mu t|)=\int^{|\mu t|}_{0}(f(x))dx$

    We also know:
    $\displaystyle \begin{align*}1&=\int_{0}^{|\mu t|}(f(x))dx + \int_{|\mu t|}^{\infty}(f(x))dx \\
    1 - \frac{1}{t}&=\int_{0}^{|\mu t|}(f(x))dx + \int_{|\mu t|}^{\infty}(f(x))dx- \frac{1}{t}\end{align*}$

    So we have to show:
    $\displaystyle \begin{align*}
    P(X<\mu t) &\geq 1 - \frac{1}{t} \\
    \int_{0}^{|\mu t|}(f(x))dx &\geq \int_{0}^{|\mu t|}(f(x))dx + \int_{|\mu t|}^{\infty}(f(x))dx- \frac{1}{t} \\
    0 &\geq \int_{|\mu t|}^{\infty}(f(x))dx- \frac{1}{t} \\
    \frac{1}{t} &\geq \int_{|\mu t|}^{\infty}(f(x))dx
    \end{align*}$

    This looks similar to what we had in the formula for $\displaystyle \mu$ except for the extra x in the integrand, so I assume this is where that comes in, but I'm not sure how to relate the two.
    Does anyone know where I should go from here? Or am I taking a wrong turn?
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  2. #2
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    Re: Continuous Random Variable Proof

    Quote Originally Posted by Greenmango View Post
    Question: If X is a random variable for which $\displaystyle P(X \geq 0)=0$ and $\displaystyle |\mu_x|<\infty$, show that $\displaystyle P(X<\mu t) \geq 1 - \frac{1}{t}$ for every $\displaystyle t \geq 1$.

    My TA told me to use the following information:
    $\displaystyle \mu=\int_{-\infty}^{\infty}(x(f(x))dx$
    Since $\displaystyle P(X \geq 0)=0$:
    $\displaystyle \mu=\int_{0}^{\infty}(x(f(x))dx$
    don't you mean

    $\displaystyle \mu=\int_{-\infty}^{0}(x(f(x))dx$
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  3. #3
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    Re: Continuous Random Variable Proof

    Oh sorry, I actually made a typo in the question, it should say P(X <= 0)
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    Re: Continuous Random Variable Proof

    Quote Originally Posted by Greenmango View Post
    Oh sorry, I actually made a typo in the question, it should say P(X <= 0)
    What is 'IT'
    Really the whole problem does not make any sense.
    Are you given a $PDF$ for the random variable $X~?$
    Or you given more about the mean $\mu~?$
    This seems to me to be a very strange question.
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    Re: Continuous Random Variable Proof

    "It" as in the question at the top marked by "Question:" as well as all the places where I make reference to the question by saying P(X >= 0). I know it's a weird question, that's why I'm having trouble haha, but it's written word for word how my professor gave it to me, no PDF and no mean!
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  6. #6
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    Re: Continuous Random Variable Proof

    Take a look at the Chebyshev Inequality
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    Re: Continuous Random Variable Proof

    Quote Originally Posted by Greenmango View Post
    Question: If X is a random variable for which $\displaystyle P(X \geq 0)=0$ and $\displaystyle |\mu_x|<\infty$, show that $\displaystyle P(X<\mu t) \geq 1 - \frac{1}{t}$ for every $\displaystyle t \geq 1$.

    My TA told me to use the following information:
    $\displaystyle \mu=\int_{-\infty}^{\infty}(x(f(x))dx$
    Since $\displaystyle P(X \geq 0)=0$:
    $\displaystyle \mu=\int_{0}^{\infty}(x(f(x))dx$
    $\displaystyle \mu=\int_{0}^{|\mu t|}(x(f(x))dx + \int_{|\mu t|}^{\infty}(x(f(x))dx$

    My own work:
    We know that:
    $\displaystyle P(X<\mu t)=\int^{\mu t}_{-\infty}(f(x))dx$
    Since $\displaystyle P(X \geq 0)=0$:
    $\displaystyle P(X<|\mu t|)=\int^{|\mu t|}_{0}(f(x))dx$
    Oh come on. How are we suppose to know what $f(x)$ is?
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  8. #8
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    Re: Continuous Random Variable Proof

    Quote Originally Posted by Plato View Post
    Oh come on. How are we suppose to know what $f(x)$ is?
    it's the probability density function of the random variable $X$

    I don't think we are supposed to know anything about it other than $X$ has a finite mean and support on the positive real numbers.
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