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Math Help - Obtain the maximum likelihood estmate of the average arrival rate

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    Obtain the maximum likelihood estmate of the average arrival rate

    Question : It is assumed that the arrival of number of calls X per hour follows Poisson distribution with parameter \lambda . A random sample X_1 = x_1, X_2 = x_2,.....X_n = x_n is taken. Obtain the maximum likelihood estimate of the average arrival rate
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    Quote Originally Posted by zorro View Post
    Question : It is assumed that the arrival of number of calls X per hour follows Poisson distribution with parameter \lambda . A random sample X_1 = x_1, X_2 = x_2,.....X_n = x_n is taken. Obtain the maximum likelihood estimate of the average arrival rate
    Do you know what sort of distribution the waiting time follows? That is your starting point .... Go back and look it up in your class notes.
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    This is what u mean

    Quote Originally Posted by mr fantastic View Post
    Do you know what sort of distribution the waiting time follows? That is your starting point .... Go back and look it up in your class notes.

    Do u mean exponential distribution????
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    Quote Originally Posted by zorro View Post
    Do u mean exponential distribution????
    Yes, but I realise now that I misread the question - this distribution is not relevant (which is not to say that it's not relevant to you in the broader context of developing and consolidating your general understanding).

    Here is a Christmas present:

    The likelihood function is L(x_1, x_2, .... x_n) = \frac{e^{-\lambda} \lambda^{x_1}}{x_1!} \cdot \frac{e^{-\lambda} \lambda^{x_2}}{x_2!} \cdot .... \frac{e^{-\lambda} \lambda^{x_n}}{x_n!} = e^{-n \lambda} \left( \frac{\lambda^{x_1 + x_2 + .... + x_n}}{x_1! x_2! .... x_n!} \right).

    Therefore \ln L = -n \lambda + (x_1 + x_2 + .... + x_n) \ln \lambda - \ln (x_1! x_2! .... x_n!).

    Your job is to solve \frac{d \ln L}{d \lambda} = 0 for \lambda. The answer is not surprising.
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