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Derive the MGF of binomial distribution and hence finds it's mean and variance.
written 5.4 years ago by
teamques10
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modified 4.1 years ago
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prashantsaini
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written 5.4 years ago by
teamques10
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MGF for binomial distribution
By defination, MGF about origin is
$ M_o(t) = E (e^{tx}) = \sum{p_ie^{txi}}$
$ = \sum{{{^h}_c}_xp^xq^{n-x}e^{tx}}$
$ = \sum{{{^h}_c}_xq^{n-x}(pe^t)^x}$
$ M_o(t) = (q + pe^t)^n -------------1$
To find mean, differentiate $ M_o(t)$ and put t = 0
$E(X) = \frac{d}{dt} [M_o(t)]_{t=0} =\frac{d}{dt} [(q + pe^t)^n]_{t=0} $ …
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written 5.4 years ago by
teamques10
★ 65k
|
MGF for binomial distribution
By defination, MGF about origin is
$ M_o(t) = E (e^{tx}) = \sum{p_ie^{txi}}$
$ = \sum{{{^h}_c}_xp^xq^{n-x}e^{tx}}$
$ = \sum{{{^h}_c}_xq^{n-x}(pe^t)^x}$
$ M_o(t) = (q + pe^t)^n -------------1$
To find mean, differentiate $ M_o(t)$ and put t = 0
$E(X) = \frac{d}{dt} [M_o(t)]_{t=0} =\frac{d}{dt} [(q + pe^t)^n]_{t=0} $ …
and 3 others joined a min ago.
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