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Math Help - 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,

    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:

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

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

    with\ \lambda=3\ and\ k=2: P_X(k)=0.22404" alt="\ 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,

    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:

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

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

    with\ \lambda=3\ and\ k=2: \ 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 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 EX=\lambda and \sigma^2=\lambda.


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

    And from Markov's inequality with m=1 we get 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 \frac{\lambda}{7}

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

    so \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 \frac{\lambda}{7}

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

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

    which would then make m=2 the better estimate
    I think you meant 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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