statistics

Mgf of bernoulli distribution proof

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The_Don. answered · 16/04/2021

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MGF of Bernoulli Distribution Proof

The Bernoulli Distribution can be expressed as Illustrated Here

$M_{x} (t) = \sum_{x = 0}^{1} e^{t x} p (x = x)$

$= \sum_{x = 0}^{1} e^{t x} p^{x} {(1 - p)}^{1 - x}$

Replacing x with 0 and 1

$= e^{0} p^{0} {(1 - p)}^{1 - 0} + e^{t} p^{1} {(1 - p)}^{1 - 1}$

$= (1 - p) + e^{t} p$

Thus the MGF of Bernoulli Distribution will be given by;

$= (1 - p) + e^{t} p$

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