{"id":2993,"date":"2019-06-28T14:40:37","date_gmt":"2019-06-28T14:40:37","guid":{"rendered":"https:\/\/analystprep.com\/study-notes\/?p=2993"},"modified":"2026-01-30T18:06:21","modified_gmt":"2026-01-30T18:06:21","slug":"define-probability-generating-functions-and-moment-generating-functions","status":"publish","type":"post","link":"https:\/\/analystprep.com\/study-notes\/actuarial-exams\/soa\/p-probability\/univariate-random-variables\/define-probability-generating-functions-and-moment-generating-functions\/","title":{"rendered":"Probability Generating Functions and Moment Generating Functions"},"content":{"rendered":"

Probability Generating Function<\/strong><\/h2>\n

The probability generating function of a discrete random variable<\/strong> is a power series representation of the random variable\u2019s probability density function as shown in the formula below:<\/p>\n

$$ \\begin{align*} \\text{G}\\left(\\text{n}\\right)&=\\text{P}\\ \\left(\\text{X}\\ =\\ 0\\right)\\bullet \\ \\text{n}^0\\ +\\ \\text{P}\\ \\left(\\text{X}\\ =\\ 1\\right)\\bullet\\ \\text{n}^1\\ +\\ \\text{P}\\ \\left(\\text{X}\\ =\\ 2\\right)\\bullet\\ \\text{n}^2\\ \\\\ &+\\ \\text{P}\\ \\left(\\text{X}\\ =\\ 3\\right)\\bullet \\ \\text{n}^3\\ +\\ \\text{P}\\ \\left(\\text{X}=4\\right)\\bullet \\ \\text{n}^4+\\cdots \\\\ &=\\sum_{\\text{i}=0}^{\\infty}{\\text{P}\\left(\\text{X}=\\text{x}_\\text{i}\\right).\\text{n}^\\text{i}}=\\text{E}\\left(\\text{n}^\\text{i}\\right) \\end{align*} $$<\/p>\n

Note: \\( G\\left( 1 \\right) =P\\left( X=0 \\right) +P\\left( X=1 \\right) +P\\left( X=2 \\right) +P\\left( X=3 \\right) +P\\left( X=4 \\right) +\\cdots P\\left( X=r \\right) =1 \\)<\/p>\n

Example: Probability Generating Function of a Discrete Random Variable<\/strong><\/h4>\n

Given the experiment of rolling a single die, find the probability generating function.<\/p>\n

Solution<\/strong><\/p>\n

We know that:<\/p>\n

$$ \\begin{align*} \\text{G}\\ \\left(\\text{n}\\right)&=\\text{E}\\left(\\text{n}^\\text{i}\\right)\\\\ &=0\\times \\text{n}^0+\\left(\\frac{1}{6}\\right)\\times\\ \\text{n}^1+\\left(\\frac{1}{6}\\right)\\times\\ \\text{n}^2+\\left(\\frac{1}{6}\\right)\\times\\ \\text{n}^3+\\left(\\frac{1}{6}\\right)\\times\\ \\text{n}^4+\\left(\\frac{1}{6}\\right)\\times\\ \\text{n}^5 \\\\ &+\\left(\\frac{1}{6}\\right)\\times\\ \\text{n}^6 \\end{align*} $$<\/p>\n

It can be useful to express \\(E(X)\\) and \\(Var(X)\\) as a function of the probability generating function as shown below:<\/p>\n

$$ \\begin{align*} E\\left( X \\right) &=G^{ \\prime }\\left( 1 \\right) \\\\ \\text{Var}\\left(\\text{X}\\right)&=\\text{G}^{\\prime \\prime}\\left(1\\right)+\\text{G}^{\\prime}\\left(1\\right)-\\left[\\text{G}^{\\prime}\\left(1\\right)\\right]^2 \\\\ \\end{align*} $$<\/p>\n

Example: Probability Generating Function of a Discrete Random Variable<\/strong><\/h4>\n

Given the experiment of rolling a single die, find \\(E(X)\\) and \\(Var(X)\\) using the probability generating function.<\/p>\n

$$ \\begin{align*} & G\\left( n \\right) =0\\bullet n^{ 0 }+\\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 1 }+\\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 2 }+\\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 3 }+\\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 4 }+\\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 5 }+\\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 6 } \\\\ & G^{ \\prime }\\left( n \\right) =\\left( \\frac { 1 }{ 6 } \\right) +2\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n+3\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 2 }+4\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 3 }+5\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 4 }+6\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 5 } \\\\ & G^{ \\prime }\\left( 1 \\right) =\\left( \\frac { 1 }{ 6 } \\right) +2\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1+3\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 2 }+4\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 3 }+5\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 4 }+6\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 5 }=3.5 \\\\ & G^{ \\prime \\prime }\\left( n \\right) =2\\bullet \\left( \\frac { 1 }{ 6 } \\right) +3\\bullet 2\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n+4\\bullet 3\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 2 }+5\\bullet 4\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 3 }+6\\bullet 5\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet n^{ 4 } \\\\ & G^{ \\prime \\prime }\\left( 1 \\right) =2\\bullet \\left( \\frac { 1 }{ 6 } \\right) +3\\bullet 2\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1+4\\bullet 3\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 2 }+5\\bullet 4\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 3 }+6\\bullet 5\\bullet \\left( \\frac { 1 }{ 6 } \\right) \\bullet 1^{ 4 } = 11.667 \\\\ & Var\\left( X \\right) =G^{ ” }\\left( 1 \\right) +G^{ ‘ }\\left( 1 \\right) -\\left[ G^{ ‘ }\\left( 1 \\right) \\right] ^{ 2 }=11.667+3.5-3.5^{ 2 }=2.92 \\\\ \\end{align*} $$<\/p>\n

Moment Generating Function<\/h3>\n

A moment generating function \\(M(t)\\) of a random variable \\(X\\) is defined for all real value of \\(t\\) by:<\/p>\n

$$ \\text{M}\\left( \\text{t} \\right) =\\text{E}\\left( { \\text{e} }^{ \\text{tX} } \\right) =\\begin{cases} \\sum _{ \\text{x} }^{ }{ { \\text{e} }^{ \\text{tX} } } \\text{p}\\left( \\text{x} \\right) , \\text{ if X is a discrete with mass function } \\text{p}\\left(\\text{x}\\right) \\\\ \\int _{ -\\infty }^{ \\infty }{ { \\text{e} }^{ \\text{tX} } } \\text{f}\\left( \\text{x} \\right) \\text{dx}, \\text{ if X is continous with density function } \\text{f}\\left(\\text{x}\\right) \\end{cases} $$<\/p>\n

Example: Moment Generating Function of a Discrete Random Variable<\/h4>\n

Given the experiment of rolling a single die, find the moment generating function.<\/p>\n

Solution<\/strong><\/p>\n

We know that:<\/p>\n

\\(M\\left( t \\right) =e^{ t1 }P\\left( X=1 \\right) +e^{ t2 }P\\left( X=2 \\right) +e^{ t3 }P\\left( X=3 \\right) +e^{ t4 }P\\left( X=4 \\right) +e^{ t5 }P\\left( X=5 \\right) +e^{ t6 }P\\left( X=6 \\right) \\)<\/p>\n

\\( M\\left( t \\right) =e^{ t }\\left( \\frac { 1 }{ 6 } \\right) +e^{ 2t }\\left( \\frac { 1 }{ 6 } \\right) +e^{ 3t }\\left( \\frac { 1 }{ 6 } \\right) +e^{ 4t }\\left( \\frac { 1 }{ 6 } \\right) +e^{ 5t }\\left( \\frac { 1 }{ 6 } \\right) +e^{ 6t }\\left( \\frac { 1 }{ 6 } \\right) \\)<\/p>\n

As with the probability generating function, it can also be useful to use the moment generating function to calculate \\(E(X)\\) and \\(Var(X)\\) as shown in the formulas below:<\/p>\n

$$ \\begin{align*} & E \\left(X \\right)= \\mu =M^{\\prime} \\left(0 \\right) \\\\ & E \\left(X^2 \\right)=M^{\\prime\\prime}\\left(0 \\right) \\\\ & Var \\left(X \\right)= {\\sigma}^2= M^{\\prime\\prime} \\left(0 \\right)-\\left[M^{‘} \\left(0 \\right) \\right]^2 \\\\ \\end{align*} $$<\/p>\n

Consider the moment generating function above. We wish to calculate \\(\\text{E}\\left(\\text{X}\\right)\\) and \\(\\text{Var}\\left(\\text{X}\\right)\\).<\/p>\n

We know that:<\/p>\n

$$ \\text{E}\\left(\\text{X}\\right)=\\mu=\\text{M}\\prime(0) $$<\/p>\n

The moment generating function is given by:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\text{e}^\\text{t}\\left(\\frac{1}{6}\\right)+\\text{e}^{2\\text{t}}\\left(\\frac{1}{6}\\right)+\\text{e}^{3\\text{t}}\\left(\\frac{1}{6}\\right)+\\text{e}^{4\\text{t}}\\left(\\frac{1}{6}\\right)+\\text{e}^{5\\text{t}}\\left(\\frac{1}{6}\\right)+\\text{e}^{6\\text{t}}\\left(\\frac{1}{6}\\right) $$<\/p>\n

Taking the first derivative, we have:<\/p>\n

$$ \\text{M}^\\prime\\left(\\text{t}\\right)=\\left(\\frac{1}{6}\\right)\\text{e}^\\text{t}+\\left(\\frac{2}{6}\\right)\\text{e}^{2\\text{t}}+\\left(\\frac{3}{6}\\right)\\text{e}^{3\\text{t}}+\\left(\\frac{4}{6}\\right)\\text{e}^{4\\text{t}}+\\left(\\frac{5}{6}\\right)\\text{e}^{5\\text{t}}+\\left(\\frac{6}{6}\\right)\\text{e}^{6\\text{t}} $$<\/p>\n

Substituting \\(t=0\\) in the above equation, we have:<\/p>\n

$$ \\text{M}^\\prime\\left(0\\right)=\\left(\\frac{1}{6}\\right)+\\left(\\frac{2}{6}\\right)+\\left(\\frac{3}{6}\\right)+\\left(\\frac{4}{6}\\right)+\\left(\\frac{5}{6}\\right)+\\left(\\frac{6}{6}\\right)=3.5 $$<\/p>\n

For the variance, we know that:<\/p>\n

$$ \\text{Var}\\left(\\text{X}\\right)=\\sigma^2=\\text{E}\\left(\\text{X}^2\\right)-\\left[\\text{E}\\left(\\text{X}\\right)\\right]^2=\\text{M}^{\\prime\\prime}\\left(0\\right)-\\left[\\text{M}^\\prime\\left(0\\right)\\right]^2 $$<\/p>\n

Now taking the second derivative of the moment generating function, we have:<\/p>\n

$$ \\begin{align*} \\text{M}^{\\prime \\prime}\\left(\\text{t}\\right)&=\\left(\\frac{1}{6}\\right)\\text{e}^\\text{t}+\\left(\\frac{4}{6}\\right)\\text{e}^{2\\text{t}}+\\left(\\frac{9}{6}\\right)\\text{e}^{3\\text{t}}+\\left(\\frac{16}{6}\\right)\\text{e}^{4\\text{t}}+\\left(\\frac{25}{6}\\right)\\text{e}^{5\\text{t}}+\\left(\\frac{36}{6}\\right)\\text{e}^{6\\text{t}} \\\\ \\Rightarrow\\ \\text{M}^\\prime\\left(0\\right)&=\\left(\\frac{1}{6}\\right)+\\left(\\frac{4}{6}\\right)+\\left(\\frac{9}{6}\\right)+\\left(\\frac{16}{6}\\right)+\\left(\\frac{25}{6}\\right)+\\left(\\frac{36}{6}\\right)=15.167 \\\\ \\text{Var}\\left(\\text{X}\\right)&=\\sigma^2=\\text{M}^{\\prime\\prime}\\left(0\\right)-\\left[\\text{M}^\\prime\\left(0\\right)\\right]^2=15.167-{3.5}^2=2.92 \\end{align*} $$<\/p>\n

Example: Moment Generating Function of Discrete Random Variables<\/h4>\n

Suppose that the discrete random variable \\(X\\) has a distribution:<\/p>\n

$$ \\text{f}\\left(\\text{x}\\right)= \\begin{cases} \\frac { 1 }{ 2 } ,\\text{x}=1 \\\\ \\frac { 3 }{ 8 } ,\\text{x}=2 \\\\ \\frac { 1 }{ 8 } ,\\text{x}=3 \\end{cases} $$<\/p>\n

a. Find the moment generating function \\(M(t)\\).<\/p>\n

Solution<\/strong><\/p>\n

We know that:$$ \\begin{align*} \\text{M}\\left(\\text{t}\\right)&=\\text{E}\\left(\\text{e}^{\\text{xt}}\\right)=\\left(\\text{x}+\\text{a}\\right)^\\text{n}=\\sum_{\\text{x}}{\\text{e}^{\\text{xt}}\\text{f}(\\text{x})}\\ \\\\ &=\\text{e}^\\text{tf}\\left(1\\right)+\\text{e}^{2\\text{t}}\\text{f}\\left(2\\right)+\\text{e}^{3\\text{t}}\\text{f}\\left(3\\right) \\\\ &=\\left(\\frac{1}{2}\\right)\\text{e}^{\\prime \\text{t}}+\\left(\\frac{3}{8}\\right)\\text{e}^{2\\text{t}}+\\left(\\frac{1}{8}\\right)\\text{e}^{3t} \\\\ &\\therefore\\ \\text{M}\\left(\\text{t}\\right)=\\frac{1}{2}\\text{e}^\\text{t}+\\frac{3}{8}\\text{e}^{2\\text{t}}+\\frac{1}{8}\\text{e}^{3\\text{t}} \\end{align*} $$<\/p>\n

b. Use the moment generating function to find the mean of X.<\/p>\n

Solution<\/strong><\/p>\n

The mean of X is \\( \\text{E}\\left(\\text{X}\\right)=\\text{M}\\prime(0) \\).We are given that:$$ \\begin{align*} \\text{M}\\left(\\text{t}\\right)&=\\frac{1}{2}\\text{e}^\\text{t}+\\frac{3}{8}\\text{e}^{2\\text{t}}+\\frac{1}{8}\\text{e}^{3t} \\\\ &{\\Rightarrow \\text{M}}^\\prime(\\text{t})=\\frac{1}{2}\\text{e}^t+\\frac{2\\times3}{8}\\text{e}^{2\\text{t}}+\\frac{3\\times1}{8}\\text{e}^{3t} \\\\ &=\\frac{1}{2}\\text{e}^t+\\frac{6}{8}\\text{e}^{2\\text{t}}+\\frac{3}{8}\\text{e}^{3\\text{t}} \\\\ \\text{M}^\\prime\\left(0\\right)&=\\frac{1}{2}\\text{e}^0+\\frac{6}{8}\\text{e}^{2\\times0}+\\frac{3}{8}\\text{e}^{3\\times0}=\\frac{1}{2}+\\frac{6}{8}+\\frac{3}{8}=\\frac{13}{8} \\end{align*} $$Mean of \\( \\text{X}= \\text{M\u2019} (0) = \\frac{13}{8} \\)<\/p>\n

c. Use the moment generating function to find the variance of X.<\/p>\n

Solution<\/strong><\/p>\n

We know that:$$ \\text{Var}\\ \\left(\\text{X}\\right) \\text{E}\\left(\\text{X}^2\\right)-\\text{E}\\left(\\text{X}\\right)^2=\\text{M}^{\\prime\\prime}\\left(0\\right)-{[\\text{M}}^{\\prime}\\left(0\\right)]2 $$From (b), we have:$$ \\begin{align*} \\text{M}^{\\prime}(\\text{t})&=\\frac{1}{2}e^t+\\frac{6}{8}\\text{e}^{2\\text{t}}+\\frac{3}{8}\\text{e}^{3\\text{t}} \\\\ {\\Rightarrow \\text{M}}^{\\prime\\prime}\\left(\\text{t}\\right)&=\\frac{1}{2}\\text{e}^\\text{t}+\\frac{2\\times6}{8}\\text{e}^{2\\text{t}}+\\frac{3\\times3}{8}\\text{e}^{3\\text{t}}=\\frac{1}{2}\\text{e}^\\text{t}+\\frac{12}{8}\\text{e}^{2\\text{t}}+\\frac{9}{8}\\text{e}^{3\\text{t}} \\\\ &{\\therefore \\text{M}}^{\\prime\\prime}\\left(0\\right)=\\frac{1}{2}\\text{e}^0+\\frac{12}{8}\\text{e}^{2\\times0}+\\frac{9}{8}\\text{e}^{3\\times0} \\\\ &=\\frac{1}{2}+\\frac{12}{8}+\\frac{9}{8}=\\frac{25}{8} \\\\ \\text{Var}\\ (\\text{X})&=\\text{M}^{\\prime\\prime}\\left(0\\right)-\\text{M}^{\\prime}(0)\u00b2 \\\\ &=\\frac{25}{8}-\\left(\\frac{13}{8}\\right)^2=\\frac{25}{8}-\\frac{169}{64}=\\frac{31}{64} \\end{align*} $$<\/p>\n

Example: Moment Generating Function of a Continuous Distribution<\/h4>\n

Given the following probability density function of a continuous random variable:<\/p>\n

$$ f\\left( x \\right) =\\begin{cases} 0.2{ e }^{ -0.2x }, & 0\\le x\\le \\infty \\\\ 0, & otherwise \\end{cases} $$<\/p>\n

Find the moment generating function.<\/p>\n

Solution<\/strong><\/p>\n

For a continuous distribution,<\/p>\n

$$ \\text{M}\\left(t\\right)=\\ \\int_{-\\infty}^{\\infty}{\\text{e}^{\\text{tx}}\\text{f}\\left(\\text{x}\\right)\\text{dx}} $$<\/p>\n

So, in this case:<\/p>\n

$$ \\begin{align*} \\text{M}\\left(\\text{t}\\right)&=\\int_{0}^{\\infty}{\\text{e}^{\\text{tx}}\\times\\left(0.2\\text{e}^{-0.2\\text{x}}\\right)\\times \\text{dx}} \\\\ \\text{M}\\left(\\text{t}\\right)&=\\int_{0}^{\\infty}{0.2\\text{e}^{\\text{x}\\left(\\text{t}-0.2\\right)}\\text{dx}=\\left[\\frac{0.2\\text{e}^{\\text{x}\\left(t-0.2\\right)}}{\\text{t}-0.2}\\right]_{\\text{x}=0}^{\\text{x}=\\infty}=-\\frac{0.2}{t-0.2}} \\end{align*} $$<\/p>\n

As with the moment generating function of the discrete distribution, we can use the moment generating function of a continuous distribution to calculate \\(E(X)\\) and \\(Var(X)\\) using the formulae below:<\/p>\n

$$ \\begin{align*} E \\left(X \\right) & = \\mu=M^{\\prime}\\left(0 \\right) \\\\ E \\left(X^2 \\right) & =M^{\\prime\\prime} \\left(0 \\right) \\\\ Var \\left(X \\right) & = {\\sigma}^2= M^{\\prime\\prime} \\left(0 \\right)-\\left[M^{\\prime} \\left(0\\right) \\right]^2 \\\\ \\end{align*} $$<\/p>\n

Example: Calculating Expectation and Variance for MGF <\/strong><\/h4>\n

Given the moment generating function shown below, calculate \\(E(X)\\) and \\(Var(X)\\).<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\frac{0.2}{\\text{t}-0.2\\ } $$<\/p>\n

Solution<\/strong><\/p>\n

Note that we can write:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\frac{0.2}{t-0.2\\ }=-0.2\\left(\\text{t}-0.2\\right)^{-1} $$<\/p>\n

So that:<\/p>\n

$$ \\text{M}^\\prime\\left(\\text{t}\\right)=0.2\\left(\\text{t}-0.2\\right)^{-2} $$<\/p>\n

We know that:<\/p>\n

$$ \\text{E}\\left(\\text{X}\\right)=\\text{M}^\\prime\\left(0\\right)=0.2\\left(-0.2\\right)^2=\\frac{1}{0.2}=5 $$<\/p>\n

Also,<\/p>\n

$$ \\begin{align*} {\\text{E}\\left(\\text{X}^2\\right)}&=\\text{M}^{\\prime\\prime} \\left(\\text{t}\\right)=-2\\times0.2\\left(\\text{t}-0.2\\right)^{-3}=-0.4\\left(\\text{t}-0.2\\right)^{-3} \\\\ \\Rightarrow \\text{M}^{\\prime\\prime}\\left(0\\right)&=-0.4\\left(-0.2\\right)^{-3}=50 \\\\ \\therefore \\text{Var}\\left(\\text{X}\\right)&=\\sigma^2=\\text{M}^{\\prime\\prime}\\left(0\\right)-\\left[\\text{M}^\\prime\\left(0\\right)\\right]^2=50-5^2=25 \\end{align*} $$<\/p>\n

Moment Generating Functions of Common Distributions<\/strong><\/h2>\n

Binomial Distribution<\/strong><\/h3>\n

The moment generating function for \\(X\\) with a binomial distribution is an alternate way of determining the mean and variance.<\/p>\n

Let us perform n<\/em> independent Bernoulli trials, each of which has a probability of success \\(p\\) and probability of failure \\(1-p\\). Thus, the probability mass function (PMF) is:<\/p>\n

$$ \\text{f}\\left(\\text{x}\\right)=\\binom{\\text{n}}{\\text{x}}\\text{p}^\\text{x}\\left(1-\\text{p}\\right)^{\\text{n}-\\text{x}} $$<\/p>\n

Where, \\(\\binom{\\text{n}}{\\text{x}}\\) can be written as \\(\\text{C}(\\text{n},\\text{x}) \\) and denotes the number of combinations of n<\/em> elements taken \\(x\\) at a time; where \\(x\\) can take on values \\(0,1,2,3, \u2026,n\\).<\/p>\n

Using the PMF, we can obtain the moment generating function of \\(X\\):<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\sum_{\\text{x}=0}^{\\text{n}}{\\text{e}^{\\text{tx}}\\binom{\\text{n}}{\\text{x}}\\text{p}^\\text{x}\\left(1-\\text{p}\\right)^{\\text{n}-\\text{x}}} $$<\/p>\n

Combine the terms with exponential of \\(x\\):<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\sum_{\\text{x}=0}^{\\text{n}}{\\left(\\text{pe}^\\text{t}\\right)^\\text{x}\\binom{\\text{n}}{\\text{x}}\\ \\left(1-\\text{p}\\right)^{\\text{n}-\\text{x}}} $$<\/p>\n

By using the binomial formula, the expression above is simply:<\/p>\n

$$ M \\left(t \\right)={\\left(p{e}^{t}+1-p \\right)}^n $$<\/p>\n

Mean and Variance<\/strong><\/h4>\n

In order to calculate the mean and variance, we need to find both \\(\\text{M}\\prime(0)\\) and \\(\\text{M}\\prime\\prime(0)\\).<\/p>\n

First, you start by calculating the derivatives, then evaluate each of them at \\(t=0\\). The 1st derivative of the moment generating function is:<\/p>\n

$$ \\text{M}^\\prime\\left(\\text{t}\\right)=\\text{n}\\left(\\text{pe}^\\text{t}+1-\\text{p}\\right)^{\\text{n}-1}\\text{pe}^\\text{t} \\\\ \\Rightarrow \\text{M}^\\prime\\left(0\\right)=\\text{n}\\left(\\text{p}+1-\\text{p}\\right)^{\\text{n}-1}\\text{p}=\\text{np} $$<\/p>\n

If we differentiate the second time, we get:<\/p>\n

$$ \\text{M}^{\\prime\\prime}\\left(\\text{t}\\right)=\\text{n}\\left(\\text{n}-1\\right)\\left(\\text{pe}^t+1-\\text{p}\\right)^{\\text{n}-2}\\left(\\text{pe}^\\text{t}\\right)^2+\\text{n}\\left(\\text{pe}^\\text{t}+1-\\text{p}\\right)^{\\text{n}-1}\\text{pe}^\\text{t} $$<\/p>\n

So that:<\/p>\n

$$ \\text{E}\\left(\\text{X}^2\\right)=\\text{M}^{\\prime\\prime}\\left(0\\right)=\\text{n}\\left(\\text{n}-1\\right)\\text{p}^2+\\text{np} $$<\/p>\n

We know that:<\/p>\n

$$ \\text{Var}\\left(\\text{X}\\right)=\\text{E}\\left(\\text{X}^2\\right)-\\left[\\text{E}\\left(\\text{X}\\right)\\right]^2=\\text{n}\\left(\\text{n}-1\\right)\\text{p}^2+\\text{np}-\\text{n}^2\\text{p}^2=\\text{np}(1-\\text{p}) $$<\/p>\n

Negative Binomial Distribution<\/strong><\/h3>\n

The moment generating function of a negative binomial distribution is given by:<\/p>\n

$$ M\\left( t \\right) ={ \\left[ \\frac { p{ e }^{ t } }{ 1-\\left( 1-p \\right) { e }^{ t } } \\right] }^{ r } $$<\/p>\n

Which is derived as follows:<\/p>\n

$$ \\begin{align*} \\text{M}\\left(\\text{t}\\right)=\\text{E}\\left[\\text{e}^{\\text{tX}}\\right]&=\\sum_{\\text{k}=\\text{r}}^{\\infty}{\\text{e}^{\\text{tk}}\\binom{\\text{k}-1}{\\text{r}-1}\\left(1-\\text{p}\\right)^{\\text{k}-\\text{r}}\\text{p}^\\text{r}} \\\\ &=\\sum_{\\text{k}=\\text{r}}^{\\infty}{\\text{e}^{\\text{tk}}\\binom{\\text{k}-1}{\\text{r}-1}\\left(1-\\text{p}\\right)^{\\text{k}-\\text{r}}\\text{p}^\\text{r}\\times\\frac{\\text{e}^{\\text{tr}}}{\\text{e}^{\\text{tr}}}} \\\\ &=\\sum_{\\text{k}=\\text{r}}^{\\infty}{\\text{e}^{\\text{t}\\left(\\text{k}-\\text{r}\\right)}\\binom{\\text{k}-1}{\\text{r}-1}\\left(1-\\text{p}\\right)^{\\text{k}-\\text{r}}{(\\text{e}}^\\text{t}{\\text{p})}^\\text{r}} \\\\ &=\\left(\\text{e}^\\text{t} \\text{p}\\right)^\\text{r}\\sum_{\\text{k}=\\text{r}}^{\\infty}{\\binom{\\text{k}-1}{\\text{r}-1}{(\\text{e}}^\\text{t}{\\left(1-\\text{p}\\right))}^{\\text{k}-\\text{r}}} \\\\ \\end{align*} $$<\/p>\n

Setting \\(j=k-r\\),<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)={\\text{e}^\\text{t}}{\\text{p}}^\\text{r}\\sum_{\\text{j}=0}^{\\infty}{{\\text{j}+\\text{r}-1}{\\text{r}-1}(\\text{e}^\\text{t}{\\left(1-\\text{p}\\right))}^\\text{j}} \\\\ \\frac{\\text{e}^\\text{t}{\\text{p})}^\\text{r}}{(1-\\text{e}^\\text{t}(1-{\\text{p}))}^\\text{r}}=\\text{M}\\left(\\text{t}\\right)=\\left[\\frac{\\text{pe}^\\text{t}}{1-\\left(1-\\text{p}\\right)\\text{e}^\\text{t}}\\right]^\\text{r} $$<\/p>\n

Geometric Distribution<\/strong><\/h3>\n

The moment generating function of geometric distribution is given by:<\/p>\n

$$ M\\left( t \\right) =\\frac { p{ e }^{ t } }{ 1-\\left( 1-p \\right) { e }^{ t } } $$<\/p>\n

The moment generating function for \\( \\text{X}\\sim geometric \\left(\\text{p}\\right)\\) is derived as:<\/p>\n

$$ \\begin{align*} \\text{M}\\left(\\text{t}\\right)&=\\text{E}[\\text{e}^{\\text{tX}}] \\\\ &=\\sum_{\\text{x}=0}^{\\infty}{\\text{e}^{\\text{tx}}\\text{p}\\left(1-\\text{p}\\right)^\\text{x}} \\\\ \\text{Step}\\ (\\text{i})&=\\text{p}\\sum_{\\text{x}=0}^{\\infty}{(\\text{e}^\\text{t}{\\left(1-\\text{p}\\right))}^\\text{x}} \\end{align*} $$<\/p>\n

Where the series in step (i) converges only if:<\/p>\n

$$ \\left(1-\\text{p}\\right)\\text{e}^\\text{t}$$<\/p>\n

Meaning, only if:<\/p>\n

$$ \\text{e}^\\text{t}<\\frac{1}{(1-\\text{p})} $$<\/p>\n

The condition becomes \\( \\text{t}<-\\ln(1-\\text{p}) \\) by taking the natural log of both sides.<\/p>\n

$$ \\text{M}(\\text{t})=\\frac{\\text{p}}{1-\\left(1-\\text{p}\\right)\\text{e}^\\text{t}} $$<\/p>\n

Poisson Distribution<\/strong><\/h3>\n

Let \\(X\\) be a discrete random variable with a Poisson distribution with parameter \\( \\lambda \\) for some \\( \\lambda\\epsilon\\text{R}>0 \\).<\/p>\n

Then the moment generating function \\({\\text{M} }_{\\text{X}}\\) of X is given by:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\text{e}^{\\lambda\\left(\\text{e}^\\text{t}-1\\right)} $$<\/p>\n

Proof<\/h4>\n

From Poisson distribution definition, \\(X\\) has a PMF:<\/p>\n

$$ \\text{P}\\left(\\text{X}=\\text{n}\\right)=\\frac{\\lambda^\\text{n} \\text{e}^{-\\lambda}}{\\text{n}!} $$<\/p>\n

From the moment generating function definition:<\/p>\n

$$ \\text{M}_\\text{X}\\left(\\text{t}\\right)=\\text{E}\\left(\\text{e}^{\\text{tX}}\\right)=\\sum_{\\text{n}=0}^{\\infty}{\\text{P}\\left(\\text{X}=\\text{n}\\right)\\text{e}^{\\text{tn}}} $$<\/p>\n

Thus,<\/p>\n

$$ \\text{M}_\\text{X}\\left(\\text{t}\\right)=\\sum_{\\text{n}=0}^{\\infty}\\frac{\\lambda^\\text{n} \\text{e}^{-\\lambda}}{\\text{n}!}\\text{e}^{\\text{tn}}\\\\ =\\text{e}^ {-\\lambda}\\sum_{\\text{n}=0}^{\\infty}\\frac{\\left(\\lambda \\text{e}^\\text{t}\\right)^\\text{n}}{\\text{n}!} $$<\/p>\n

By the power series expansion for the exponential function:<\/p>\n

$$ =\\text{e}^{-\\lambda}\\sum_{\\text{n}=0}^{\\infty}\\frac{\\left(\\lambda \\text{e}^\\text{t}\\right)^\\text{n}}{\\text{n}!}{=\\text{e}}^{-\\lambda}\\text{e}^{\\lambda \\text{e}^\\text{t}\\ \\ \\ \\ \\ }\\ \\ =\\text{e}^{\\lambda\\left(\\text{e}^\\text{t}-1\\right)} $$<\/p>\n

Uniform Distribution<\/h3>\n

Let \\(\\text{X}~\\text{U}[\\text{a}\\ .\\ .\\text{b}] \\) for some \\(\\text{a},\\ \\text{b}\\ \\epsilon\\mathbb{\\text{R}},\\ \\text{a}\\neq\\ \\text{b},\\) where U is the continuous uniform distribution<\/p>\n

Then the moment generating function of \\(X\\), \\({\\text{M}}_{\\text{X}}\\), is given by:<\/p>\n

$$ {\\text{M}}_{\\text{X}}\\left(\\text{t}\\right) = \\begin{cases} \\frac { \\text{e}^{ \\text{tb} }-\\text{e}^{ \\text{ta} } }{ \\text{t}(\\text{b}-\\text{a}) } & \\text{t}\\neq 0 \\\\ 1 & \\text{t}= 0 \\\\ \\end{cases} $$<\/p>\n

Proof<\/h4>\n

The definition of continuous uniform distribution, \\(X\\) has a PMF of:<\/p>\n

$$ \\text f_{\\text X}\\left(\\text x\\right)= \\begin{cases} \\frac { 1 }{ \\text{b}-\\text{a} } & \\text{a}\\le \\text{x}\\le \\text{b} \\\\ 0 & \\text{otherwise} \\\\ \\end{cases} $$<\/p>\n

The moment generating function definition goes as follows:<\/p>\n

$$ \\text{M}_\\text{X}\\left(t\\right)=\\mathbb{\\text{E}}\\left[\\text{e}^{\\text{tX}}\\right]=\\int_{\\infty}^{\\infty}{\\text{e}^{\\text{tx}}\\text{f}_\\text{X}(\\text{x})\\text{dx}} $$<\/p>\n

Then, we first have to consider that \\(\\text{t}\\neq 0\\).<\/p>\n

Thus,<\/p>\n

$$ \\begin{align*} \\text{M}_\\text{X}\\left(\\text{t}\\right)&=\\int_{\\infty}^{\\text{a}}{0\\text{e}^{\\text{tx}}\\text{dx}+\\int_{\\text{a}}^{\\text{b}}{\\frac{\\text{e}^{\\text{tx}}}{\\text{b}-\\text{a}}\\text{dx}+\\int_{\\text{b}}^{\\infty}{0\\text{e}^{\\text{tx}}\\text{dx}}}} \\\\ &{=\\left[\\frac{\\text{e}^{\\text{tx}}}{\\text{t}\\left(\\text{b}-\\text{a}\\right)}\\right]}_\\text{a}^\\text{b}\\ \\text{ by primitive of e}^{\\text{ax}}\\ (\\text{a fundamental theorem of calculus}) \\\\ \\text{M}_\\text{X}(\\text{t})&=\\frac{\\text{e}^{\\text{tb}}-\\text{e}^{\\text{ta}}}{\\text{t}(\\text{b}-\\text{a})} \\end{align*} $$<\/p>\n

Note<\/strong>: For \\(t=0\\), we have \\( \\mathbb{\\text{E}}\\left[\\text{X}^0\\right]=\\mathbb{\\text{E}}\\left[1\\right]=1 \\).<\/p>\n

Exponential Distribution<\/h3>\n

If \\(X\\) has an exponential distribution, then the formulae below apply:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\frac{\\lambda}{\\lambda-\\text{t}} $$<\/p>\n

Proof<\/h4>\n

$$ \\begin{align*} \\text{M}\\left(\\text{t}\\right)&=\\text{E}\\left(\\text{e}^{\\text{tX}}\\right)\\\\ &=\\int_{0}^{\\infty}{\\text{e}^{\\text{tx}}\\lambda \\text{e}^{-\\lambda \\text{x}}\\text{dx}} \\\\ &=\\lambda\\int_{0}^{\\infty}{\\text{e}^{-\\left(\\lambda-t\\right)\\text{x}}\\ \\text{dx}\\ }\\\\ &=\\frac{\\lambda}{\\lambda-\\text{t}}\\ for\\ \\text{t}<\\lambda \\end{align*} $$<\/p>\n

Note that:<\/p>\n

$$ \\begin{align*} \\text{M}^\\prime\\left(\\text{t}\\right)& =\\frac{\\lambda}{\\left(\\lambda-\\text{t}\\right)^2} \\\\ \\Rightarrow \\text{M}^\\prime\\left(0\\right) & =\\frac{1}{\\lambda} \\\\ \\therefore \\text{E}\\left(\\text{X}\\right) & =\\frac{1}{\\lambda} \\end{align*} $$<\/p>\n

Also,<\/p>\n

$$ \\begin{align*} \\text{M}^{\\prime\\prime}\\left(\\text{t}\\right)&=\\frac{2\\lambda}{\\left(\\lambda-t\\right)^3}\\\\ \\Rightarrow\\ \\text{M}^{\\prime\\prime}\\left(0\\right) & =\\frac{2}{\\lambda^2} \\\\ \\therefore\\ \\text{Var}\\left(\\text{X}\\right)& =\\frac{2}{\\lambda^2}-\\left(\\frac{1}{\\lambda}\\right)^2=\\frac{1}{\\lambda^2} \\end{align*} $$<\/p>\n

Gamma Distribution<\/h3>\n

If \\(X\\) has a gamma distribution over the interval \\( [0,\\ \\infty] \\) with parameters \\(k\\) and \\({ \\lambda } \\), then the formulae below apply:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\left[\\frac{\\lambda}{\\lambda-\\text{t}}\\right]^\\text{s} $$<\/p>\n

Normal Distribution<\/h3>\n

Let \\( \\text{x}\\sim \\text{N}(\\mu,\\ \\sigma^2) \\), then the moment generating function is:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\text{e}^{\\mu \\text{t}+\\frac{\\left(\\sigma^2\\text{t}^2\\right)}{2}} $$<\/p>\n

Intuitively, the MGF of a standard normal variable is:<\/p>\n

$$ \\text{M}\\left(\\text{t}\\right)=\\text{e}^\\frac{\\text{t}^2}{2} $$<\/p>\n

Learning Outcome<\/i><\/p>\n

Topic 2.e: Univariate Random Variables – Define probability generating functions and moment generating functions and use them to calculate probabilities and moments.<\/i><\/p>\n","protected":false},"excerpt":{"rendered":"

Probability Generating Function The probability generating function of a discrete random variable is a power series representation of the random variable\u2019s probability density function as shown in the formula below: $$ \\begin{align*} \\text{G}\\left(\\text{n}\\right)&=\\text{P}\\ \\left(\\text{X}\\ =\\ 0\\right)\\bullet \\ \\text{n}^0\\ +\\ \\text{P}\\…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[98],"tags":[],"class_list":["post-2993","post","type-post","status-publish","format-standard","hentry","category-univariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"\nProbability & Moment Generating Functions | SOA P<\/title>\n<meta name=\"description\" content=\"Learn probability and moment generating functions, compare PGF vs MGF, and see how they are used to derive moments in 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