Given \\(X\\) and \\(Y\\) are independent random variables, then the probability density function of \\(a=X+Y\\) can be shown by the equation below:<\/p>\r\n
$$ { f }_{ X+Y }\\left( a \\right) =\\int _{ -\\infty }^{ \\infty }{ { f }_{ X }\\left( a-y \\right) } { f }_{ Y }\\left( y \\right) dy $$<\/p>\r\n
The cumulative distribution function, also called the convolution<\/strong> of \\(X\\) and \\(Y\\), can be shown by the equation below:<\/p>\r\n $$ { F }_{ X+Y }\\left( a \\right) =\\int _{ -\\infty }^{ \\infty }{ { F }_{ X }\\left( a-y \\right) } { f }_{ Y }\\left( y \\right) dy $$<\/p>\r\n Given two independent uniform random variables shown by the probability density functions below, find the probability density function of \\(a=X+Y\\).<\/p>\r\n $$ f\\left( x \\right) =\\begin{cases} \\frac { 1 }{ 2 } & 0\\le X\\le 2 \\\\ 0 & \\text{otherwise} \\end{cases} \\quad f\\left( y \\right) =\\begin{cases} \\frac { 1 }{ 2 } & 0\\le Y\\le 2 \\\\ 0 & \\text{otherwise} \\end{cases}$$<\/p>\r\n Solution<\/strong><\/p>\r\n We know that:<\/p>\r\n $$ { f }_{ X+Y }\\left( a \\right) =\\int _{ -\\infty }^{ \\infty }{ { f }_{ X }\\left( a-y \\right) } \\left( \\cfrac{ 1 }{ 2 } \\right) dy $$<\/p>\r\n Therefore,<\/p>\r\n $$\\begin{align} { f }_{ X+Y }\\left( a \\right) &=\\int _{ -\\infty }^{ \\infty }{\u00a0 \\left( \\cfrac{ 1 }{ 2 }\\right) . \\left( \\cfrac{ 1 }{ 2 } \\right)} dy\\\\ &=\\int _{ -\\infty }^{ \\infty }{\\frac{1}{4}} dy\u00a0 \\end{align}$$<\/p>\r\n We now need to find the interval of \\(a=X+Y\\). Clearly, X varies in the interval [0,2] and Y varies in the interval [0,2], and thus \\(a=X+Y\\) must vary in the interval [0,4]. However, note that from the intervals of X and Y, we have two possible intervals for \\(a=X+Y\\): [0,2) and (2,4].<\/p>\r\n Now considering the first interval [0,2), we will integrate \\({ f }_{ X+Y }\\left( a \\right)\\) from 0 to \\(a\\) since he lower interval of Y is 0 and that \\(Y\\le a\\). Thus for this case:<\/p>\r\n $$ { f }_{ X+Y }\\left( a \\right) =\\int _{ 0 }^{ a }{ \\left( \\frac { 1 }{ 2 } \\right) } \\left( \\cfrac{ 1 }{ 2 } \\right) dy=\\cfrac { 1 }{ 4 } \\text{ a for } 0 < a < 2 $$<\/p>\r\n For the interval (2,4], we will integrate\u00a0 \\({ f }_{ X+Y }\\left( a \\right)\\) from \\(a-2\\) to 2 since, the upper bound of Y is 2 and for the lower bound, note that \\(Y\\le a\\) and that \\(0\\le Y\\le 2\\). Thus:<\/p>\r\n $$ \\begin{align}{ f }_{ X+Y }\\left( a \\right) &=\\int _{ a-2 }^{ 2 }{ \\left( \\frac { 1 }{ 2 } \\right) } \\left( \\frac { 1 }{ 2 } \\right) dy\\\\ &={ \\left[ \\frac { 1 }{ 4 } y \\right] }_{ y=a-2 }^{ y=2 }\\\\&=\\left( \\frac { 1 }{ 4 } \\right) \\left( 2-\\left( a-2 \\right) \\right)\\\\& =\\left( \\frac { 1 }{ 4 } \\right) \\left( 4-a \\right) \\text{ for } 2 < a < 4 \\end{align}$$<\/p>\r\n Therefore:<\/p>\r\n $${ f }_{ X+Y }\\left( a \\right) =\\begin{cases} \\frac { 1 }{ 4 } & 0\\le a\\le 2 \\\\ \\left( \\frac { 1 }{ 4 } \\right) \\left( 4-a \\right) & 2\\le a\\le 4 \\\\ 0 & \\text{otherwise} \\end{cases}$$<\/p>\r\n If \\(X\\) and \\(Y\\) are independent random variables, then the following are true:<\/p>\r\n $$ P \\left(X=x \\text{ and } Y=y \\right) = P\\left(X=x \\right) \\bullet P\\left(Y=y \\right)=f \\left(x \\right)\\bullet f \\left(y \\right) $$<\/p>\r\n Given the experiment of rolling two dice simultaneously, find the probability of rolling a 6 on both dice.<\/p>\r\n $$ P \\left(X=6 \\text{ and } Y=6 \\right) = P \\left(X=6 \\right) \\bullet P\\left(Y=6 \\right) = \\cfrac {1}{6} \\bullet \\cfrac {1}{6} = \\cfrac {1}{36} $$<\/p>\r\n $$ E \\left(XY \\right)=E \\left(X \\right) \\bullet E\\left(Y \\right) $$<\/p>\r\n Given two independent uniform random variables shown by the probability density function below, find \\(E(X+Y)\\) and \\(Var (X+Y)\\).<\/p>\r\n $$ Solution<\/strong><\/p>\r\n We know that for two independent random variables:<\/p>\r\n $$ Now,<\/p>\r\n $$ And<\/p>\r\n $$ For the variance, we know that:<\/p>\r\n $$ We have that:<\/p>\r\n $$ And<\/p>\r\n $$ \\begin{align*} We can also use the probability density function of \\(X+Y\\) that was derived above to verify this solution:<\/p>\r\n $$ \\begin{align*} It is also true that for two independent random variables,<\/p>\r\n $$ Given the following moment generating functions of independent random variables, \\(X\\) and \\(Y\\), find the moment generating function of \\(X+Y\\).<\/p>\r\n $$ { M }_{ X }\\left( t \\right) =exp\\left\\{ .2\\left( { e }^{ t }-1 \\right) \\right\\} \\\\ \\text{And,} \\\\ $$ { M }_{ X+Y }\\left( t \\right) = exp\\left\\{ .2\\left( { e }^{ t }-1 \\right) \\right\\} \\bullet exp\\left\\{ .3\\left( { e }^{ t }-1 \\right) \\right\\} $$<\/p>\r\n Solution<\/strong><\/p>\r\n We know that for two independent random variables:<\/p>\r\n $$ Thus, in this case:<\/p>\r\n $$ \\begin{align*} If \\(X\\) and \\(Y\\) are independent Poisson random variables with parameters \\(\\lambda_x \\) and \\(\\lambda_y\\) respectively, then \\({ {X}+ {Y}}\\) is a Poison distribution with parameter \\(\\lambda=\\lambda_ {x}+\\lambda_ {y} \\).<\/p>\r\n Prove that the sum of two Poisson variables also follows a Poisson distribution.<\/p>\r\n Solution<\/strong><\/p>\r\n The probability generating function (PGF) of a discrete random variable \\(x\\) is given by:<\/p>\r\n $$ {G}\\left( {t}\\right)=\\sum_{\\left(all\\ {x}\\right)}{ {P}\\left( {X}= {x}\\right) {t}^ {x}} $$<\/p>\r\n Consider \\( {X} \\sim {Po}(\\lambda) \\).<\/p>\r\n Where \\( {P}\\left( {X}= {x}\\right)=\\frac{\\lambda^ {x} {e}^{-\\lambda}}{x!} \\).<\/p>\r\n $$ \\begin{align*} Let \\( {Y} \\sim {Po}(\\mu) \\)<\/p>\r\n $$ {G}_ {Y}\\left( {t}\\right)= {e}^{-\\mu\\left(1-{t}\\right)} $$<\/p>\r\n Thus,<\/p>\r\n $$ \\begin{align*} Hence \\( {X}+ {Y}\\sim {P}(\\lambda+\\mu) \\).<\/p>\r\n Let \\(X\\), \\(Y\\), and \\(Z\\) be independent Poisson random variables with \\( {E}\\left( {X}\\right)=1,\\ {E}\\left( {Y}\\right)=2 \\text{ and } {E}\\left( {Z}\\right)=3\\).<\/p>\r\n Find \\({P}( {X}+ {Y}+ {Z}\\le2) \\).<\/p>\r\n Solution<\/strong><\/p>\r\n Fact 1:<\/p>\r\n The Poisson PMF is given by:<\/p>\r\n $$ \\begin{align*} Fact 2:<\/p>\r\n If \\( {X}\\sim \\text{ Poisson } (\\lambda_1), \\)<\/p>\r\n and \\( {Y}\\sim \\text{ Poisson } \\left(\\lambda_2\\right), {\\text{ X and Y iind.}}, \\)<\/p>\r\n then \\( {X}+ {Y}\\sim \\text{ Poisson }(\\lambda_1+\\lambda_2) \\).<\/p>\r\n So,<\/p>\r\n $$ \\begin{align*} If \\(X\\) and \\(Y\\) are independent normally distributed random variables with parameters \\(\\mu_ {x}, \\sigma_ {x} \\) and \\( \\mu_ {y}, \\sigma_ {y} \\) respectively, then \\(X+Y\\) is normally distributed with parameters \\(\\mu=\\mu_ {x}+\\mu_ {y}\\) and \\(\\sigma^2=\\sigma_ {x}^2+\\sigma_ {y}^2\\).<\/p>\r\n Suppose that \\(X\\) is a normal random variable with a given mean and variance so that:<\/p>\r\n $$ {f}_ {X}\\left( {x}\\right)=\\frac{1}{\\sqrt{(2\\pi\\sigma_ {x})}} {e}^\\frac{{-\\left( {x}-\\mu_ {x}\\right)}^2}{2\\sigma_ {x}^2} $$<\/p>\r\n Similarly, \\(Y\\) is normal with a given mean and variance so that:<\/p>\r\n $$ {f}_ {Y}\\left( {y}\\right)=\\frac{1}{\\sqrt{(2\\pi\\sigma_ {y})}} {e}^\\frac{{-\\left( {y}-\\mu_ {y}\\right)}^2}{2\\sigma_ {y}^2} $$<\/p>\r\n Assume that \\(X\\) and \\(Y\\) are independent. We wish to find the sum of the two normal random variables by deriving the PDF of \\(Z\\):<\/p>\r\n $$ {Z}= {X}+ {Y} $$<\/p>\r\n We apply the convolution formulae:<\/p>\r\n Where we have \\( {X}\\sim {N}\\left(\\mu_ {x},\\ \\sigma_ {x}^2\\right)\\text{ and } {Y}\\sim {N}(\\mu_ {y},\\ \\sigma_ {y}^2) \\) stated as:<\/p>\r\n $$ {f}_ {X}\\left( {x}\\right)=\\frac{1}{\\sqrt{(2\\pi\\sigma_ {x})}} {e}^\\frac{{-\\left( {x}-\\mu_ {x}\\right)}^2}{2\\sigma_ {x}^2}, {f}_ {Y}\\left( {y}\\right)=\\frac{1}{\\sqrt{(2\\pi\\sigma_ {y})}} {e}^\\frac{{-\\left( {y}-\\mu_ {y}\\right)}^2}{2\\sigma_ {y}^2} $$<\/p>\r\n Using the convolution formulae \\( {f}_ {Z}\\left( {z}\\right)=\\int_{-\\infty}^{\\infty}{ {f}_ {X}\\left( {x}\\right) {f}_ {Y}( {z}- {x}) {dx}} \\), we plug in the form for the density of \\(X\\) at \\( {f}_ {X}( {x}) \\) and form for the density of \\(Y\\) at \\( {f}_ {Y}( {Z}- {X}) \\).<\/p>\r\n Therefore,<\/p>\r\n $$ \\begin{align*} Note:<\/strong><\/p>\r\n The argument above is based on the sum of two independent normal random variables. Suppose we have the sum of three normally independent random variables such that \\(X+Y+W\\).<\/p>\r\n From the above discussion, \\( {X}+ {Y} \\) is normal, \\(W\\) is assumed to be normal. We also assume that \\( {X}+ {Y} \\) and \\(W\\) are all independent.<\/p>\r\n Therefore, \\( {X}+ {Y} \\) are independent from \\(W\\), so we are dealing with the sum of two independent normal random variables. In that case, the sum of \\( {X}+ {Y}+ {W} \\) is also going to be normal.<\/p>\r\n Learning Outcome<\/strong><\/i><\/p>\r\n Topic 2.f: Univariate Random Variables – Determine the sum of independent random variables (Poisson and normal).<\/strong><\/i><\/p>\r\n","protected":false},"excerpt":{"rendered":" The Sum of Independent Random Variables Given \\(X\\) and \\(Y\\) are independent random variables, then the probability density function of \\(a=X+Y\\) can be shown by the equation below: $$ { f }_{ X+Y }\\left( a \\right) =\\int _{ -\\infty }^{…<\/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-2995","post","type-post","status-publish","format-standard","hentry","category-univariate-random-variables","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"\nExample: Sum of Independent Random Variables<\/h4>\r\n
Example: Sum of Independent Random Variables<\/h4>\r\n
Example: Expectation of Two Independent Random Variables<\/h4>\r\n
\r\n\\text{f}\\left(\\text{x}\\right)=\\begin{cases} \\frac { 1 }{ 2 } ,& 0\\le \\text{X}\\le 2 \\\\ 0 & \\text{otherwise} \\\\ \\end{cases};\\ \\ \\ \\ \\text{f}\\left(\\text{y}\\right)=\\begin{cases} \\frac { 1 }{ 2 } ,& 0\\le \\text{X}\\le 2 \\\\ 0 & \\text{otherwise} \\\\ \\end{cases}
\r\n$$<\/p>\r\n
\r\n{E}\\left( {X}+ {Y}\\right)= {E}\\left( {X}\\right)+ {E}\\left( {Y}\\right)
\r\n$$<\/p>\r\n
\r\n{E}\\left( {X}\\right)=\\frac{2+0}{2}=1
\r\n$$<\/p>\r\n
\r\n{E}\\left( {Y}\\right)=\\frac{2+0}{2}=1 \\\\
\r\n\\Rightarrow\\ {E}\\left( {X}+ {Y}\\right)=1+1=2
\r\n$$<\/p>\r\n
\r\n{Var}\\left( {X}+ {Y}\\right)= {Var}\\left(
\r\n{X}\\right)+ {Var}( {Y})
\r\n$$<\/p>\r\n
\r\n{Var}\\left( {X}\\right)=\\frac{\\left(2-0\\right)^2}{12}=\\frac{1}{3}
\r\n$$<\/p>\r\n
\r\n{Var}\\left(Y\\right) & =\\frac{\\left(2-0\\right)^2}{12}=\\frac{1}{3} \\\\
\r\n\\Rightarrow\\ {Var}\\left( {X}+ {Y}\\right) & =\\frac{1}{3}+\\frac{1}{3}=\\frac{2}{3}
\r\n\\end{align*} $$<\/p>\r\n
\r\n{f}_\\left( {X}+ {Y}\\right)\\left( {x}\\right)&=\\begin{cases} \\frac { 1 }{ 4 } {x},& 0\\le {x}\\le 2 \\\\ \\left( \\frac { 1 }{ 4 } \\right) \\left( 4- {x} \\right) ,& 2\\le {x}\\le 4 \\\\ 0 & \\text{otherwise} \\\\ \\end{cases} \\\\
\r\n{E}\\left( {E}+ {Y}\\right)&=\\int_{0}^{2}{\\left(\\frac{1}{4}\\right) {x}. {x}\\ {dx}\\ +\\int_{2}^{4}{\\left(\\frac{1}{4}\\right)\\left(4- {x}\\right) {x}\\ {dx}}} \\\\
\r\n&=\\left[\\frac{1}{12} {x}^3\\right]_{ {x}=0}^{ {x}=2}+\\left[\\left(\\frac{1}{2}\\right) {x}^2-\\frac{1}{12} {x}^3\\right]_{ {x}=2}^{ {x}=4}=\\frac{2}{3}+\\frac{4}{3}=\\frac{6}{3}=2\\
\r\n\\end{align*} $$<\/p>\r\n
\r\n{M}_{ {X}+ {Y}}\\left( {t}\\right)= {M}_ {X} {t}\\times\\ {M}_ {Y} {t}
\r\n$$<\/p>\r\nExample: Moment Generating Function of the Two Independent Random Variables<\/h4>\r\n
\r\n\\quad { M }_{ Y }\\left( t \\right) =exp\\left\\{ .3\\left( { e }^{ t }-1 \\right) \\right\\} $$<\/p>\r\n
\r\n{M}_{ {X}+ {Y}}\\left( {t}\\right)= {M}_ {X} {t}\\times\\ {M}_ {Y} {t}
\r\n$$<\/p>\r\n
\r\n{M}_{ {X}+ {Y}}\\left( {t}\\right)&= {e}^{0.2\\left( {e}^ {t}-1\\right)}\\times\\ {e}^{0.3\\left( {e}^ {t}-1\\right)} \\\\
\r\n{M}_{ {X}+ {Y}}\\left( {t}\\right)&= {e}^{(0.2 {e}^ {t}-0.2+0.3 {e}^ {t}-0.3)} \\\\
\r\n&= {e}^{0.5 {e}^ {t}-0.5}= {e}^{0.5\\left( {e}^ {t}-1\\right)}
\r\n\\end{align*} $$<\/p>\r\nSum of Poisson Random Variables<\/h3>\r\n
Example: Sum of Poisson Random Variables<\/h4>\r\n
\r\n{G}_ {X}\\left( {t}\\right)&= {E}( {t}^ {x}) \\\\
\r\n&=\\sum_{ {x}=0}^{\\infty}{ {t}^ {x}\\ \\frac{\\lambda^ {x} {e}^{-\\lambda}}{ {x}!}} \\\\
\r\n&= {e}^{-\\lambda}\\sum_{ {x}=0}^{\\infty}\\frac{\\left( {t}\\lambda\\right)^ {x}}{ {x}!} \\\\
\r\n&{= {e}}^{-\\lambda} {e}^{ {t}\\lambda} \\\\
\r\n&{= {e}}^{-\\lambda+ {t}\\lambda} \\\\
\r\n{ {G}_ {X}( {t})}&= {e}^{-\\lambda\\left(1- {t}\\right)}
\r\n{G}_{ {X}+ {Y}}\\left( {t}\\right)&= {G}_ {X}\\left( {t}\\right)+ {G}_ {Y}( {t}) \\\\
\r\n&{= {e}}^{-\\lambda\\left(1- {t}\\right)} {e}^{-\\mu\\left(1- {t}\\right)} \\\\
\r\n&{= {e}}^{-(\\lambda+\\mu)\\left(1- {t}\\right)}
\r\n\\end{align*} $$<\/p>\r\nExample: Sum of the Poisson Random Variables<\/h4>\r\n
\r\n& {P}\\left( {X}= {x}\\right)=\\frac{ {e}^{-\\lambda}\\lambda^ {x}}{ {x}!},\\ \\ \\ {x}=0,1,2,3,\\ldots \\\\
\r\n& {E}\\left( {X}\\right)=\\lambda
\r\n\\end{align*} $$<\/p>\r\n
\r\n& {X}+ {Y}+ {Z} \\sim Poisson\\left(\\lambda_1+\\lambda_2+\\lambda_3\\right)=1+2+3=6 \\\\
\r\n& {P}\\left( {X}+ {Y}+ {Z}\\right)\\le2 \\\\
\r\n& {P}\\left( {X}+ {Y}+ {Z}=0\\right)+ {P}\\left( {X}+ {Y}+ {Z}=1\\right)+ {P}\\left(X+Y+Z=2\\right) \\\\
\r\n&=\\frac{ {e}^{-6}6^0}{0!}+\\frac{ {e}^{-6}6^1}{1!}+\\frac{ {e}^{-6}6^2}{2!} \\\\
\r\n&{= {e}}^{-6}(1+6+18) \\\\
\r\n&=25 {e}^{-6}=0.061968804\\approx0.062
\r\n\\end{align*} $$<\/p>\r\nSum of Normal Random Variables<\/h3>\r\n
Proof:<\/h4>\r\n
\r\n{f}_ {z}( {z}) & =\\int _{ -\\infty }^{ \\infty }{ \\cfrac { 1 }{ \\sqrt { \\left( 2\\pi { \\sigma }_{ {x} } \\right) } } exp } \\left\\{ -\\cfrac { { \\left( {x}-{ \\mu }_{ {x} } \\right) }^{ 2 } }{ 2{ \\sigma }_{ {x} }^{ 2 } } \\right\\} \\cfrac { 1 }{ \\sqrt { \\left( 2\\pi { \\sigma }_{ {y} } \\right) } } exp\\left\\{ -\\cfrac { { \\left( {z}- {x}-{ \\mu }_{ {y} } \\right) }^{ 2 } }{ 2{ \\sigma }_{ {y} }^{ 2 } } \\right\\} {dx} \\\\
\r\n{f}_ {z}( {z}) & =\\cfrac { 1 }{ \\sqrt { \\left( 2\\pi \\left( { \\sigma }_{ {x} }^{ 2 } \\right) \\right) } } {exp}\\left\\{ -\\cfrac { { \\left( {z}-{ \\mu }_{ {x} }-{ \\mu }_{ {y} } \\right) }^{ 2 } }{ 2\\left( { \\sigma }_{ {x} }^{ 2 }+{ \\sigma }_{ {y} }^{ 2 } \\right) } \\right\\}
\r\n\\end{align*} $$<\/p>\r\n