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    Binomial distribution with different success probabilities




    I have n independent trials with different success probabilities θi for i=1 to i=n
    I want to show that...

    μ=nθ with θ= (1/n) ∑θi for i=1 to i=n

    I guess this makes sense intuitively for θ to be the average of all θi. Can someone help show this? The proof for a constant θ is straight forward. I'll leave the variance for another day.

    Edit: To be clear, this is not a homework exercise. I found it in my book while reviewing the material. I attached the problem below.
    Last edited by Buckeye; 07-21-2016 at 03:42 PM.

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    Re: Binomial distribution with different success probabilities

    I'm not clear what your question is exactly.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Binomial distribution with different success probabilities

    I apologize. I'm new to this website. The attached problem is courtesy of John E. Freund's Mathematical Statistics with Applications eighth edition. It's an undergraduate text. I found the problem interesting because from what I know of the binomial distribution, the success probability is held constant.
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    Re: Binomial distribution with different success probabilities


    Define X_i to be the outcome of the ith trial. Then X = \sum_{i=1}^n X_i.

    Use that to rewrite \mu_X = E[X] by inserting the definition of X into the expected value. See if you can break that up. Keep in mind that E[X_i] = \theta_i
    I don't have emotions and sometimes that makes me very sad.

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    Buckeye (07-21-2016)

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