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    Markov Inequality




    If X_1,X_2,X_3...X_{20} are independent random variables with common distribution Poisson(1). Find with markov inequality a upper limit for P(\sum_{i=1}^{20}X_i>15)

    I think there was a typing problem in this matter, because look
    I know that X = \sum_{i=1}^{20}X_i \rightarrow Poisson(20) so with markov inequality P(X\geq15)\leq\frac{E[X]}{15} = \frac{20}{15}>1

    Not make sense the probability greather than 1.

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    Re: Markov Inequality

    You inequality says the the probability is less than or equal to 1.25 - not greater than 1. This is a true statement is it not?
    I don't have emotions and sometimes that makes me very sad.

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    Re: Markov Inequality

    Quote Originally Posted by Dason View Post
    You inequality says the the probability is less than or equal to 1.25 - not greater than 1. This is a true statement is it not?
    Yes the sentence is true, but if I say that the probability is less than or equal than 1.25 it can be for example 1.1 which does not make sense to me.

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    Re: Markov Inequality

    If I tell you that .7 is less than 1.25 that doesn't mean that .7 can be 1.1. The *inequality* is just giving you a bound. All that matters is if the true probability actually falls in the range that your inequality dictates. All of the values in your inequality don't need to be possible values for the probability. Just like how even if the inequality said that the probability was < .9 you wouldn't say that it's wrong because a probability of -0.3 isn't possible (and note that -0.3 falls in the bounds of the inequality given...)
    I don't have emotions and sometimes that makes me very sad.

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    Re: Markov Inequality

    Quote Originally Posted by Dason View Post
    If I tell you that .7 is less than 1.25 that doesn't mean that .7 can be 1.1. The *inequality* is just giving you a bound. All that matters is if the true probability actually falls in the range that your inequality dictates. All of the values in your inequality don't need to be possible values for the probability. Just like how even if the inequality said that the probability was < .9 you wouldn't say that it's wrong because a probability of -0.3 isn't possible (and note that -0.3 falls in the bounds of the inequality given...)
    Yeah, now I get it.

    Question b) Use CLT for obtaining an aproximation for P(\sum_{i=1}^{20}X_i>15)
    What I do
    P(\sum_{i=1}^{20}X_i>15)=1-P(\sum_{i=1}^{20}X_i \leq15)=1-P(\frac{\sum_{i=1}^{20}X_i - \sum_{i=1}^{20}\mu_i}{\sqrt{\sum_{i=1}^{20}\sigma_i^2}}\leq\frac{15-\sum_{i=1}^{20}1)}{\sqrt{\sum_{i=1}^{20}1}}
    Now
    1-P(Z\leq-1.1)=1-(1-\Phi(1.1))=\Phi(1.1)=0.8643

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    Re: Markov Inequality

    Seems alright to me.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Markov Inequality


    Quote Originally Posted by Dason View Post
    Seems alright to me.
    Thank you. Solved.

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