Poisson Distribution

[Amit Goyal]

Of course you know what to do :-)

[Vibhuti Dhingra]

On Sun, May 27, 2012 at 4:19 PM, Amit Goyal <amit.k...@gmail.com> wrote:

Of course you know what to do :-)

 

1) Mgf = e^lambda(e^t-1)
2) Mean = Variance = lambda :)

--
Thanks & Regards,

Vibhuti Dhingra

[Srishti Gupta]

yes same as vibhuti:
.mgf
E[e^xt] = [summation e^tx* lamda^x *e^[-lamda] ] / x!
           = [e^[-lamda] summation ( (e^t) lamda)^x/x!]
           = which onsolving gives
             
 e^lambda(e^t-1)

[Srishti Gupta]

            =   e^lambda(e^t-1)

similary mean n variance are same am getting!! :)

[akg]

MGF is e^(lambda(e^t - 1))

Mean = Variance = lambda

On May 27, 5:35 pm, Srishti Gupta <sweetpriya.sris...@gmail.com>
wrote:

[Tuhin Chatterjeee]

same:)