# Thread: Obtain the maximum likelihood estmate of the average arrival rate

1. ## 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

2. Originally Posted by zorro
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.

3. ## This is what u mean

Originally Posted by mr fantastic
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????

4. Originally Posted by zorro
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.