Expectation of Bernoulli Distribution

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Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.

Then the expectation of $X$ is given by:

$\expect X = p$

Proof 1

From the definition of expectation:

$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$

By definition of Bernoulli distribution:

$\expect X = 1 \times p + 0 \times \paren {1 - p}$

Hence the result.


Proof 2

Follows directly from Expectation of Binomial Distribution, putting $n = 1$.


Proof 3

From the Probability Generating Function of Bernoulli Distribution, we have:

$\map {\Pi_X} s = q + p s$

where $q = 1 - p$.

From Expectation of Discrete Random Variable from PGF, we have:

$\expect X = \map { {\Pi_X}'} 1$

From Derivatives of PGF of Bernoulli Distribution:

$\map { {\Pi_X}'} s = p$

Hence the result.


Proof 4

From Moment Generating Function of Bernoulli Distribution, the moment generating function of $X$, $M_X$, is given by:

$\map {M_X} t = q + p e^t$

where $q = 1 - p$.

By Moment in terms of Moment Generating Function:

$\expect X = \map {M_X'} 0$

We have:

\(\ds \map {M_X'} t\) \(=\) \(\ds \frac \d {\d t} \paren {q + p e^t}\)
\(\ds \) \(=\) \(\ds p e^t\) Derivative of Constant, Derivative of Exponential Function

Setting $t = 0$ gives:

\(\ds \expect X\) \(=\) \(\ds p e^0\)
\(\ds \) \(=\) \(\ds p\) Exponential of Zero