Determine the sum of independent random variables (Poisson and normal)

Given \(X\) and \(Y\) are independent random variables, then the probability density function of \(X+Y\) can be shown by the equation below:

$$ { f }_{ X+Y }\left( a \right) =\int _{ -\infty }^{ \infty }{ { f }_{ X }\left( a-y \right) } { f }_{ Y }\left( y \right) dy $$

The cumulative distribution function, also called the convolution of \(X\) and \(Y\), can be shown by the equation below:

$$ { F }_{ X+Y }\left( a \right) =\int _{ -\infty }^{ \infty }{ { F }_{ X }\left( a-y \right) } { f }_{ Y }\left( y \right) dy $$

Example

Given two independent uniform random variables shown by the probability density functions below, find the probability density function of \(X+Y\).

$$ f\left( x \right) =\begin{cases} \frac { 1 }{ 2 } & 0\le X\le 2 \\ 0 & otherwise \end{cases} \quad f\left( y \right) =\begin{cases} \frac { 1 }{ 2 } & 0\le Y\le 2 \\ 0 & otherwise \end{cases}$$

$$ { f }_{ X+Y }\left( a \right) =\int _{ -\infty }^{ \infty }{ { f }_{ X }\left( a-y \right) } \left( { 1 }/{ 2 } \right) dy $$
Since \(f(y)\) is only defined for y between 0 and 2 then

\( { f }_{ X+Y }\left( a \right) =\int _{ 0 }^{ 2 }{ { f }_{ X }\left( a-y \right) } \left( { 1 }/{ 2 } \right) dy = \)

Also since \(f(x)\) is only defined for \(x\) between 0 and 2 then \( 0 < a-y < 2\) and

$$ \begin{align*}
& { f }_{ X+Y }\left( a \right) =\int _{ 0 }^{ a }{ \left( \frac { 1 }{ 2 } \right) } \left( { 1 }/{ 2 } \right) dy={ 1 }/{ 4 }\quad \quad \quad \quad a\quad for\quad 0 < a < 2 \\
& { f }_{ X+Y }\left( a \right) =\int _{ a-2 }^{ 2 }{ \left( \frac { 1 }{ 2 } \right) } \left( \frac { 1 }{ 2 } \right) dy={ \left[ \frac { 1 }{ 4 } y \right] }_{ y=a-2 }^{ y=2 }=\left( \frac { 1 }{ 4 } \right) \left( 2-\left( a-2 \right) \right) =\left( \frac { 1 }{ 4 } \right) \left( 4-a \right) \quad for\quad 2 < a < 4 \\
\end{align*}
$$

If \(X\) and \(Y\) are independent random variables, then the following are true:

$$ P \left(X=x\quad and \quad Y=y \right) = P\left(X=x \right) \ast P\left(Y=y \right)=f \left(x \right)\ast f \left(y \right) $$

Example

Given the experiment of rolling two dice simultaneously, find the probability of rolling a 6 on both dice.

$$ P \left(X=6 \quad and \quad Y=6 \right) = P \left(X=6 \right) \ast P\left(Y=6 \right) = {1}/{6} \ast {1}/{6} = {1}/{36} $$

\( E \left(XY \right)=E \left(X \right) \ast E\left(Y \right) \)

Example

Given the experiment of rolling two dice simultaneously, find \(E(XY)\).

$$ \begin{align*}
& E\left(XY \right) = E \left(X \right) \ast E \left(Y \right) = 3.5 \ast 3.5 = 12.25 \\
& E \left(X+Y \right)=E \left(X \right)+E \left(Y \right) \\
\end{align*}
$$

\( Var \left(X+Y \right)=Var \left(X \right)+Var \left(Y \right) \)

Example

Given two independent uniform random variables shown by the probability density functions below, find \(E(X+Y)\) and \(Var(X+Y)\).

$$ f\left( x \right) =\begin{cases} \frac { 1 }{ 2 } & 0\le X\le 2 \\ 0 & otherwise \end{cases} \quad f\left( y \right) =\begin{cases} \frac { 1 }{ 2 } & 0\le Y\le 2 \\ 0 & otherwise \end{cases} $$

$$ \begin{align*}
& E \left(X+Y \right) = E \left(X \right) + E \left(Y \right) \\
& E \left(X \right) = {\left(2+0\right)}/{2} = 1 \\
& E \left(Y \right) = {\left(2+0\right)}/{2} = 1 \\
& E \left(X+Y \right) = 1 + 1 = 2 \\ \\
& Var \left (X+Y \right) = Var \left(X \right) + Var \left(Y \right) \\
& Var \left(X \right) = {\left(2-0 \right)^2}/{12} = {1}/{3} \\
& Var \left(Y \right) = {\left(2-0 \right)^2}/{12} = {1}/{3} \\
& Var \left (X+Y \right) = {1}/{3}+{1}/{3}={2}/{3} \\
\end{align*}
$$

We can also use the probability density function of \(X+Y\) that was derived above to verify this solution:

$$ { f }_{ X+Y }\left( x \right) =\begin{cases} \cfrac { 1 }{ 4 } x & 0\le x\le 2 \\ \left( \cfrac { 1 }{ 4 } \right) \left( 4-x \right) & 2\le x\le 4 \\ 0 & otherwise \end{cases} $$

\( \begin{align*}
E\left( X+Y \right) & =\int _{ 0 }^{ 2 }{ \left( \frac { 1 }{ 4 } \right) x\ast xdx } ={ \left[ \frac { 1 }{ 12 } { x }^{ 3 } \right] }_{ x=0 }^{ x=2 }=\frac { 8 }{ 12 } =\frac { 2 }{ 3 } \\
& +\int _{ 2 }^{ 4 }{ \left( \frac { 1 }{ 4 } \right) \left( 4-x \right) \ast xdx } ={ \left[ \left( \frac { 1 }{ 2 } \right) { x }^{ 2 }-\frac { 1 }{ 12 } { x }^{ 3 } \right] }_{ x=2 }^{ x=4 }=\frac { 4 }{ 3 } \\
\end{align*}
\)

\( \begin{align*}
& E \left(X+Y \right) = {2}/{3} + {4}/{3} = {6}/{3} = 2 \\
\end{align*}
\)

\( \begin{align*}
& { M }_{ X+Y }\left( t \right) ={ M }_{ X }\left( t \right) \ast { M }_{ Y }\left( t \right)
\end{align*}
\)

Example

Given the following moment generating functions of independent random variables, \(X\) and \(Y\), find the moment generating function of \(X+Y\).

$$ { M }_{ X }\left( t \right) =exp\left\{ .2\left( { e }^{ t }-1 \right) \right\} \quad \quad \quad { M }_{ Y }\left( t \right) =exp\left\{ .3\left( { e }^{ t }-1 \right) \right\} \quad \quad \quad $$

\( { M }_{ X+Y }\left( t \right) = exp\left\{ .2\left( { e }^{ t }-1 \right) \right\} \ast exp\left\{ .3\left( { e }^{ t }-1 \right) \right\} \)

\({ M }_{ X+Y }\left( t \right) =exp\left\{ .2{ e }^{ t }-.2+.3{ e }^{ t }-.3 \right\} =exp\left\{ .5{ e }^{ t }-.5 \right\} =exp\left\{ .5\left( { e }^{ t }-1 \right) \right\} \)

Sum of Normal Random Variables

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 distribution with parameters \(\mu= \mu_x+\mu_y\) and \({ \sigma }^{ 2 }={ \sigma }_{ x }^{ 2 }+{ \sigma }_{ y }^{ 2 }\).

Sum of Poisson Random Variables

If \(X\) and \(Y\) are independent Poisson random variables with parameters \(\lambda_x\) and \(\lambda_y\) respectively, then \(X+Y\) is a Poisson distribution with parameter \(\lambda = \lambda_x + \lambda_y\).

 

Learning Outcome

Topic 2.f: Univariate Random Variables – Determine the sum of independent random variables (Poisson and normal).


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