avewright/proofwiki-math huggingface dataset:
Theorem: Bernoulli Process as Geometric Distribution

Let $\sequence {X_i}$ be a binomial experiment with parameter $p$.
Let $\EE$ be the experiment which consists of performing the Bernoulli trial $X_i$ until a failure occurs, and then stop.
Let $k$ be the number of successes before a failure is encountered.
Then $k$ is modelled by a geometric distribution with parameter $p$.

Proof:
Follows directly from the definition of geometric distribution.
Let $X$ be the discrete random variable defined as the number of successes before a failure is encountered.
Thus the last trial (and the last trial only) will be a failure, and the others will be successes.
The probability that $k$ successes are followed by a failure is:
:$\map \Pr {X = k} = p^k \paren {1 - p}$
Hence the result.
{{Qed}}
Theorem: Function of Discrete Random Variable

Let $X$ be a discrete random variable on the probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $g: \R \to \R$ be any real function.
Then $Y = g \sqbrk X$, defined as:
:$\forall \omega \in \Omega: \map Y \omega = g \sqbrk {\map X \omega}$
is also a discrete random variable.

Proof:
As $\Img X$ is countable, then so is $\Img {g \sqbrk X}$.
Now consider $g^{-1} \sqbrk Y$.
We have that:
:$\forall x \in \R: \map {X^{-1} } x \in \Sigma$
We also have that:
:$\ds \forall y \in \R: \map {g^{-1} } y = \bigcup_{x: \map g x = y} \set x$
But $\Sigma$ is a sigma-algebra and therefore closed for unions.
Thus:
:$\forall y \in \R: \map {X^{-1} } {\map {g^{-1} } y} \in \Sigma$
Hence the result.
{{qed}}
Theorem: Expectation of Function of Discrete Random Variable

Let $X$ be a discrete random variable.
Let $\expect X$ be the expectation of $X$.
Let $g: \R \to \R$ be a real function.
Then:
:$\ds \expect {g \sqbrk X} = \sum_{x \mathop \in \Omega_X} \map g x \, \map \Pr {X = x}$
whenever the sum is absolutely convergent.

Proof:
Let $\Omega_X = \Img X = I$.
Let $Y = g \sqbrk X$.
Thus:
:$\Omega_Y = \Img Y = g \sqbrk I$
So:
{{begin-eqn}}
{{eqn | l = \expect Y
      | r = \sum_{y \mathop \in g \sqbrk I} y \, \map \Pr {Y = y}
      | c = 
}}
{{eqn | r = \sum_{y \mathop \in g \sqbrk I} y \sum_{ {x \mathop \in I} \atop {\map g x \mathop = y} } \map \Pr {X = x}
      | c = Probability Mass Function of Function of Discrete Random Variable
}}
{{eqn | r = \sum_{x \mathop \in I} \map g x \, \map \Pr {X = x}
      | c = 
}}
{{end-eqn}}
From the definition of expectation, this last sum applies only when the last sum is absolutely convergent.
{{qed}}
Theorem: Probability Mass Function of Function of Discrete Random Variable

Let $X$ be a discrete random variable.
Let $Y = \map g X$, where $g: \R \to \R$ is a real function.
Then the probability mass function of $Y$ is given by:
:$\ds \map {p_Y} y = \sum_{x \mathop \in \inv g y} \map \Pr {X = x}$

Proof:
By Function of Discrete Random Variable we have that $Y$ is itself a discrete random variable.
Thus:
{{begin-eqn}}
{{eqn | l = \map {p_Y} y
      | r = \map \Pr {Y = y}
      | c = 
}}
{{eqn | r = \map \Pr {\map g X = y}
      | c = 
}}
{{eqn | r = \map \Pr {X \in \inv g y}
      | c = 
}}
{{eqn | r = \sum_{x \mathop \in \inv g y} \map \Pr {X = x}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Variance as Expectation of Square minus Square of Expectation

Let $X$ be a random variable.
Then the variance of $X$ can be expressed as:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
That is, it is the expectation of the square of $X$ minus the square of the expectation of $X$.

Proof:
=== Discrete Random Variable ===
{{:Variance as Expectation of Square minus Square of Expectation/Discrete}}
Theorem: Derivative of Geometric Sequence

Let $x \in \R: \size x < 1$.
Then:
:$\ds \sum_{n \mathop \ge 1} n x^{n - 1} = \frac 1 {\paren {1 - x}^2}$

Proof:
We have from Power Rule for Derivatives that:
:$\ds \frac \d {\d x} \sum_{n \mathop \ge 1} x^n = \sum_{n \mathop \ge 1} n x^{n - 1}$
But from Sum of Infinite Geometric Sequence: Corollary:
:$\ds \sum_{n \mathop \ge 1} x^n = \frac x {1 - x}$
The result follows by Power Rule for Derivatives and the Chain Rule for Derivatives applied to $\dfrac x {1 - x}$.
{{qed}}
Theorem: Expectation of Shifted Geometric Distribution

Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac 1 p$

Proof:
From the definition of expectation:
:$\expect X = \ds \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
By definition of shifted geometric distribution:
:$\expect X = \ds \sum_{k \mathop \in \Omega_X} k p \paren {1 - p}^{k - 1}$
Let $q = 1 - p$:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = p \sum_{k \mathop \ge 0} k q^{k - 1}
      | c = as $\Omega_X = \N$
}}
{{eqn | r = p \sum_{k \mathop \ge 1} k q^{k - 1}
      | c = The term in $k = 0$ vanishes
}}
{{eqn | r = p \frac 1 {\paren {1 - q}^2}
      | c = Derivative of Geometric Sequence
}}
{{eqn | r = \frac p {p^2}
      | c = as $q = 1 - p$
}}
{{eqn | r = \frac 1 p
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Shifted Geometric Distribution

Let $X$ be a discrete random variable with the shifted geometric distribution with parameter $p$.
Then the expectation of $X$ is given by:
:$\expect X = \dfrac 1 p$

Proof:
From the Probability Generating Function of Shifted Geometric Distribution, we have:
:$\map {\Pi_X} s = \dfrac {p s} {1 - q s}$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
      | r = \map {\frac \d {\d s} } {\frac {p s} {1 - q s} }
}}
{{eqn | r = \frac p {\paren {1 - q s}^2}
      | c = Derivatives of PGF of Shifted Geometric Distribution
}}
{{end-eqn}}
Plugging in $s = 1$:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} 1
      | r = \frac p {\paren {1 - q}^2}
}}
{{eqn | r = \frac p {p^2}
      | c = as $q = 1 - p$
}}
{{eqn | r = \frac 1 p
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Expectation of Bernoulli Distribution

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:
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.
{{qed}}
Theorem: Expectation of Bernoulli Distribution

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:
Follows directly from Expectation of Binomial Distribution, putting $n = 1$.
{{qed}}
Theorem: Expectation of Bernoulli Distribution

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:
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.
{{qed}}
Theorem: Expectation of Bernoulli Distribution

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:
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: 
{{begin-eqn}}
{{eqn	| l = \map {M_X'} t
	| r = \frac \d {\d t} \paren {q + p e^t}
}}
{{eqn	| r = p e^t
	| c = Derivative of Constant, Derivative of Exponential Function
}}
{{end-eqn}}
Setting $t = 0$ gives: 
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = p e^0
}}
{{eqn	| r = p
	| c = Exponential of Zero
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Binomial Distribution

Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$

Proof:
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$
Thus:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \sum_{k \mathop = 0}^n k \binom n k p^k q^{n - k}
      | c = {{Defof|Binomial Distribution}}, with $p + q = 1$
}}
{{eqn | r = \sum_{k \mathop = 1}^n k \binom n k p^k q^{n - k}
      | c = since for $k = 0$, $k \dbinom n k p^k q^{n - k} = 0$
}}
{{eqn | r = \sum_{k \mathop = 1}^n n \binom {n - 1} {k - 1} p^k q^{n - k}
      | c = Factors of Binomial Coefficient: $k \dbinom n k = n \dbinom {n - 1} {k - 1}$
}}
{{eqn | r = n p \sum_{k \mathop = 1}^n \binom {n - 1} {k - 1} p^{k - 1} q^{\paren {n - 1} - \paren {k - 1} }
      | c = taking out $n p$ and using $\paren {n - 1} - \paren {k - 1} = n - k$
}}
{{eqn | r = n p \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j}
      | c = putting $m = n - 1, j = k - 1$
}}
{{eqn | r = n p
      | c = Binomial Theorem and $p + q = 1$
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Binomial Distribution

Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$

Proof:
From Binomial Experiment has Binomial Distribution, we see that $X$ as defined here is a sum of discrete random variables $Y_i$ that model the Bernoulli distribution:
:$\ds X = \sum_{i \mathop = 1}^n Y_i$
Each of the Bernoulli trials is independent of each other, by definition of a binomial experiment.
It follows that:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \expect {\sum_{i \mathop = 1}^n Y_i }
}}
{{eqn | r = \sum_{i \mathop = 1}^n \expect {Y_i}
      | c = Sum of Expectations of Independent Trials
}}
{{eqn | r = \sum_{i \mathop = 1}^n p
      | c = Expectation of Bernoulli Distribution
}}
{{eqn | r = n p
      | c = Sum of Identical Terms
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Binomial Distribution

Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$

Proof:
From the Probability Generating Function of Binomial Distribution, we have:
:$\map {\Pi_X} s = \paren {q + p s}^n$
where $q = 1 - p$.
From Expectation of Discrete Random Variable from PGF, we have:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}     
{{eqn | l = \map {\Pi'_X} s
      | r = \map {\frac \d {\d s} } {q + p s}^n
      | c = 
}}
{{eqn | r = n p \paren {q + p s}^{n - 1}
      | c = Derivatives of PGF of Binomial Distribution
}}
{{end-eqn}}
Plugging in $s = 1$:
:$\map {\Pi'_X} 1 = n p \paren {q + p}$
Hence the result, as $q + p = 1$.
{{qed}}
Theorem: Expectation of Binomial Distribution

Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.
Then the expectation of $X$ is given by:
:$\expect X = n p$

Proof:
From Moment Generating Function of Binomial Distribution, the moment generating function of $X$, $M_X$, is given by:
:$\ds \map {M_X} t = \paren {1 - p + p e^t}^n$
By Moment in terms of Moment Generating Function:
:$\ds \expect X = \map {M_X'} 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \map {M_X'} t 
	| r = \frac \d {\d t} \paren {1 - p + p e^t}^n
}}
{{eqn	| r = \map {\frac \d {\d t} } {1 - p + p e^t} \map {\frac \d {\map \d {1 - p + p e^t} } } {1 - p + p e^t}^n
	| c = Chain Rule for Derivatives
}}
{{eqn	| r = n p e^t \paren {1 - p + p e^t}^{n - 1}
	| c = Derivative of Exponential Function, Derivative of Power
}}
{{end-eqn}}
Setting $t = 0$ gives:
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = n p e^0 \paren {1 - p + p e^0}^{n - 1}
}}
{{eqn	| r = n p \paren {1 - p + p}^{n - 1}
	| c = Exponential of Zero
}}
{{eqn	| r = n p
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$

Proof:
From the definition of expectation:
:$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
By definition of Poisson distribution:
:$\ds \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$
Then:
{{begin-eqn}}
{{eqn | l = \expect X
      | r = \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\paren {k - 1}!} \lambda^{k-1}
      | c = as the $k = 0$ term vanishes
}}
{{eqn | r = \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}
      | c = putting $j = k - 1$
}}
{{eqn | r = \lambda e^{-\lambda} e^{\lambda}
      | c = Taylor Series Expansion for Exponential Function
}}
{{eqn | r = \lambda
}}
{{end-eqn}}
{{qed}}
Theorem: Expectation of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$

Proof:
From Probability Generating Function of Poisson Distribution:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Discrete Random Variable from PGF:
:$\expect X = \map {\Pi'_X} 1$
We have:
{{begin-eqn}}
{{eqn | l = \map {\Pi'_X} s
      | r = \frac \d {\d s} e^{-\lambda \paren {1 - s} }
      | c = 
}}
{{eqn | r = \lambda e^{- \lambda \paren {1 - s} }
      | c = Derivatives of PGF of Poisson Distribution
}}
{{end-eqn}}
Plugging in $s = 1$:
:$\map {\Pi'_X} 1 = \lambda e^{- \lambda \paren {1 - 1} } = \lambda e^0$
Hence the result from Exponential of Zero:
:$e^0 = 1$
{{qed}}
Theorem: Expectation of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ is given by:
:$\expect X = \lambda$

Proof:
From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
By Moment in terms of Moment Generating Function:
:$\expect X = \map {M_X'} 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \map {M_X'} t 
	| r = \map {\frac \d {\d t} } {e^{\lambda \paren {e^t - 1} } }
}}
{{eqn	| r = \map {\frac \d {\d t} } {\lambda \paren {e^t - 1} } \frac \d {\map \d {\lambda \paren {e^t - 1} } } \paren {e^{\lambda \paren {e^t - 1} } }
	| c = Chain Rule for Derivatives
}}
{{eqn	| r = \lambda e^t e^{\lambda \paren {e^t - 1} }
	| c = Derivative of Exponential Function
}}
{{end-eqn}}
Setting $t = 0$ gives: 
{{begin-eqn}}
{{eqn	| l = \expect X
	| r = \lambda e^0 e^{\lambda \paren {e^0 - 1} }
}}
{{eqn	| r = \lambda
	| c = Exponential of Zero
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Bernoulli Distribution

Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$

Proof:
From the definition of variance:
:$\var X = \expect {\paren {X - \expect X}^2}$
From the Expectation of Bernoulli Distribution, we have $\expect X = p$.
Then by definition of Bernoulli distribution:
{{begin-eqn}}
{{eqn | l = \expect {\paren {X - \expect X}^2}
      | r = \paren {1 - p}^2 \times p + \paren {0 - p}^2 \times \paren {1 - p}
      | c = 
}}
{{eqn | r = p - 2 p^2 + p^3 + p^2 - p^3
      | c = 
}}
{{eqn | r = p - p^2
      | c = 
}}
{{eqn | r = p \paren {1 - p}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Bernoulli Distribution

Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$

Proof:
From Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \, \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
      | r = 1^2 \times p + 0^2 \times \paren {1 - p}
      | c = 
}}
{{eqn | r = p
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \var X
      | r = \expect {X^2} - \paren {\expect X}^2
      | c = 
}}
{{eqn | r = p - p^2
      | c = Expectation of Bernoulli Distribution
}}
{{eqn | r = p \paren {1 - p}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Bernoulli Distribution

Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$

Proof:
We can simply use the Variance of Binomial Distribution, putting $n = 1$.
{{qed}}
Theorem: Variance of Bernoulli Distribution

Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$

Proof:
From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map { {\Pi_X}''} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Bernoulli Distribution, we have:
:$\map {\Pi_X} s = q + p s$
where $q = 1 - p$.
From Expectation of Bernoulli Distribution, we have $\mu = p$.
We have $\map { {\Pi_X}''} s = 0$ from Derivatives of PGF of Bernoulli Distribution.
Hence:
:$\var X = 0 - \mu - \mu^2 = p - p^2 = p \paren {1 - p}$
{{qed}}
Theorem: Variance of Bernoulli Distribution

Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
:$X \sim \Bernoulli p$
Then the variance of $X$ is given by:
:$\var X = p \paren {1 - p}$

Proof:
From Moment Generating Function of Bernoulli Distribution, the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = q + p e^t$
From Variance as Expectation of Square minus Square of Expectation, we have: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment in terms of Moment Generating Function:
:$\expect {X^2} = \map {M_X''} 0$
We have: 
{{begin-eqn}}
{{eqn	| l = \map {M_X''} t
	| r = \frac {\d^2} {\d t^2} \paren {q + p e^t}
}}
{{eqn	| r = p \frac \d {\d t} \paren {e^t}
	| c = Derivative of Constant, Derivative of Exponential Function
}}
{{eqn	| r = p e^t
	| c = Derivative of Exponential Function
}}
{{end-eqn}}
Setting $t = 0$ gives: 
{{begin-eqn}}
{{eqn	| l = \expect {X^2}
	| r = p e^0
}}
{{eqn	| r = p
	| c = Exponential of Zero
}}
{{end-eqn}}
In Expectation of Bernoulli Distribution, it is shown that: 
:$\expect X = p$
So: 
{{begin-eqn}}
{{eqn	| l = \var X
	| r = p - p^2
}}
{{eqn	| r = p \paren {1 - p}
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$

Proof:
From the definition of Variance as Expectation of Square minus Square of Expectation:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
:$\ds \expect {X^2} = \sum_{x \mathop \in \Omega_X} x^2 \, \map \Pr {X = x}$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
      | r = \sum_{k \mathop \ge 0} {k^2 \dfrac 1 {k!} \lambda^k e^{-\lambda} }
      | c = {{Defof|Poisson Distribution}}
}}
{{eqn | r = \lambda e^{-\lambda} \sum_{k \mathop \ge 1} {k \dfrac 1 {\paren {k - 1}!} \lambda^{k - 1} }
      | c = Note change of limit: term is zero when $k=0$
}}
{{eqn | r = \lambda e^{-\lambda} \paren {\sum_{k \mathop \ge 1} {\paren {k - 1} \dfrac 1 {\paren {k - 1}!} \lambda^{k - 1} } + \sum_{k \mathop \ge 1} {\frac 1 {\paren {k - 1}!} \lambda^{k - 1} } }
      | c = straightforward algebra
}}
{{eqn | r = \lambda e^{-\lambda} \paren {\lambda \sum_{k \mathop \ge 2} {\dfrac 1 {\paren {k - 2}!} \lambda^{k - 2} } + \sum_{k \mathop \ge 1} {\dfrac 1 {\paren {k - 1}!} \lambda^{k - 1} } }
      | c = Again, note change of limit: term is zero when $k-1=0$
}}
{{eqn | r = \lambda e^{-\lambda} \paren {\lambda \sum_{i \mathop \ge 0} {\dfrac 1 {i!} \lambda^i} + \sum_{j \mathop \ge 0} {\dfrac 1 {j!} \lambda^j} }
      | c = putting $i = k - 2, j = k - 1$
}}
{{eqn | r = \lambda e^{-\lambda} \paren {\lambda e^\lambda + e^\lambda}
      | c = Taylor Series Expansion for Exponential Function
}}
{{eqn | r = \lambda \paren {\lambda + 1}
      | c = 
}}
{{eqn | r = \lambda^2 + \lambda
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \var X
      | r = \expect {X^2} - \paren {\expect X}^2
      | c = 
}}
{{eqn | r = \lambda^2 + \lambda - \lambda^2
      | c = Expectation of Poisson Distribution: $\expect X = \lambda$
}}
{{eqn | r = \lambda
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Variance of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$

Proof:
From Variance of Discrete Random Variable from PGF, we have:
:$\var X = \map {\Pi' '_X} 1 + \mu - \mu^2$
where $\mu = \expect X$ is the expectation of $X$.
From the Probability Generating Function of Poisson Distribution, we have:
:$\map {\Pi_X} s = e^{-\lambda \paren {1 - s} }$
From Expectation of Poisson Distribution, we have:
:$\mu = \lambda$
From Derivatives of PGF of Poisson Distribution, we have:
:$\map {\Pi' '_X} s = \lambda^2 e^{-\lambda \paren {1 - s} }$
Putting $s = 1$ using the formula $\map {\Pi' '_X} 1 + \mu - \mu^2$:
:$\var X = \lambda^2 e^{-\lambda \paren {1 - 1} } + \lambda - \lambda^2$
and hence the result.
{{qed}}
Theorem: Variance of Poisson Distribution

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the variance of $X$ is given by:
:$\var X = \lambda$

Proof:
From Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by: 
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From Variance as Expectation of Square minus Square of Expectation, we have: 
:$\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment in terms of Moment Generating Function:
:$\expect {X^2} = \map {M_X''} 0$
In Expectation of Poisson Distribution, it is shown that: 
:$\map {M_X'} t = \lambda e^t e^{\lambda \paren {e^t - 1} }$
Then:
{{begin-eqn}}
{{eqn	| l = \map {M''_X} t
	| r = \map {\frac \d {\d t} } {\lambda e^t e^{\lambda \paren {e^t - 1} } }
}}
{{eqn	| r = \lambda \map {\frac \d {\d t} } {e^{\lambda \paren {e^t - 1} + t} }
}}
{{eqn	| r = \lambda \map {\frac \d {\d t} } {\lambda \paren {e^t - 1} + t} \frac \d {\map \d {\lambda \paren {e^t - 1} + t} } \paren {e^{\lambda \paren {e^t - 1} + t} }
	| c = Chain Rule for Derivatives
}}
{{eqn	| r = \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}
	| c = Derivative of Power, Derivative of Exponential Function
}}
{{end-eqn}}
Setting $t = 0$: 
{{begin-eqn}}
{{eqn	| l = \expect {X^2}
	| r = \lambda \paren {\lambda e^0 + 1} e^{\lambda \paren {e^0 - 1} + 0}
}}
{{eqn	| r = \lambda \paren {\lambda + 1}
	| c = Exponential of Zero
}}
{{eqn	| r = \lambda^2 + \lambda
}}
{{end-eqn}}
From Expectation of Poisson Distribution: 
:$\expect X = \lambda$
So: 
:$\var X = \lambda^2 + \lambda - \lambda^2 = \lambda$
{{qed}}