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Thread: Poisson Distribution and Markov's Inequality

  1. #1
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    Poisson Distribution and Markov's Inequality

    "The random variable X has a Poisson distribution with parameter λ = 3. By employing Markovís inequality,

    $\displaystyle P(|X|\geq\epsilon)\leq\frac{E(|X|k)}{\epsilon^k}$

    calculate bounds for P (X ≥ 7) in the cases where k = 1, 2. Which of the two estimates is better?"


    What I've done:

    $\displaystyle Using\ P_X(k)=\frac{e^{-\lambda}\lambda^k}{k!}$

    $\displaystyle with\ \lambda=3\ and\ k=1:\ P_X(k)=0.14936$

    $\displaystyle with\ \lambda=3\ and\ k=2:$$\displaystyle \ P_X(k)=0.22404$

    I'm not sure if thats right or have any idea which is a better estimate. Could someone help explain please? Thanks

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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Poisson Distribution and Markov's Inequality

    Quote Originally Posted by carla1985 View Post
    "The random variable X has a Poisson distribution with parameter λ = 3. By employing Markovís inequality,

    $\displaystyle P(|X|\geq\epsilon)\leq\frac{E(|X|k)}{\epsilon^k}$

    calculate bounds for P (X ≥ 7) in the cases where k = 1, 2. Which of the two estimates is better?"


    What I've done:

    $\displaystyle Using\ P_X(k)=\frac{e^{-\lambda}\lambda^k}{k!}$

    $\displaystyle with\ \lambda=3\ and\ k=1:\ P_X(k)=0.14936$

    $\displaystyle with\ \lambda=3\ and\ k=2:$$\displaystyle \ P_X(k)=0.22404$

    I'm not sure if thats right or have any idea which is a better estimate. Could someone help explain please? Thanks
    Hi carla1985!

    You seem to be mixing up the symbol k.
    Let's disambiguate by not using k.

    Let's write Markov's inequality as $\displaystyle P(X\geq a)\leq\frac{E(X^m)}{a^m}$.We'll look at cases where m=1,2.

    Furthermore, note that a Poisson distribution has $\displaystyle EX=\lambda$ and $\displaystyle \sigma^2=\lambda$.


    Then from the PDF of the Poisson distribution, we know that: $\displaystyle P(X \ge 7) \ge P(X=7) = \frac{e^{-\lambda}\lambda^7}{7!}$.

    And from Markov's inequality with m=1 we get $\displaystyle P(X\geq 7)\leq\frac{E(X^1)}{7^1} = \frac{\lambda}{7}$.
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    Re: Poisson Distribution and Markov's Inequality

    Ah ok, so with m=1 $\displaystyle \frac{\lambda}{7}$

    so with m=2 $\displaystyle EX=\lambda+\lambda^2$

    so $\displaystyle \frac{\lambda+\lambda^2}{7^2}=\frac{12}{49}$?

    which would then make m=2 the better estimate
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    Super Member ILikeSerena's Avatar
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    Re: Poisson Distribution and Markov's Inequality

    Quote Originally Posted by carla1985 View Post
    Ah ok, so with m=1 $\displaystyle \frac{\lambda}{7}$

    so with m=2 $\displaystyle EX=\lambda+\lambda^2$

    so $\displaystyle \frac{\lambda+\lambda^2}{7^2}=\frac{12}{49}$?

    which would then make m=2 the better estimate
    I think you meant $\displaystyle E(X^2)=\lambda+\lambda^2$.

    But yes!
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  5. #5
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    Re: Poisson Distribution and Markov's Inequality

    yes thats what i meant. thank you very much
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