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

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

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:

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

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: Sum of Independent Random Variables

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

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

Solution

We know that:

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

Therefore,

$$\begin{align} { f }_{ X+Y }\left( a \right) &=\int _{ -\infty }^{ \infty }{  \left( \cfrac{ 1 }{ 2 }\right) . \left( \cfrac{ 1 }{ 2 } \right)} dy\\ &=\int _{ -\infty }^{ \infty }{\frac{1}{4}} dy  \end{align}$$

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].

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:

$$ { f }_{ X+Y }\left( a \right) =\int _{ 0 }^{ a }{ \left( \frac { 1 }{ 2 } \right) } \left( \cfrac{ 1 }{ 2 } \right) dy=\cfrac { 1 }{ 4 } \text{ a for } 0 < a < 2 $$

For the interval (2,4], we will integrate  \({ f }_{ X+Y }\left( a \right)\) 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:

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

Therefore:

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

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

$$ P \left(X=x \text{ and } Y=y \right) = P\left(X=x \right) \bullet P\left(Y=y \right)=f \left(x \right)\bullet f \left(y \right) $$

Example: Sum of Independent Random Variables

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

$$ P \left(X=6 \text{ and } Y=6 \right) = P \left(X=6 \right) \bullet P\left(Y=6 \right) = \cfrac {1}{6} \bullet \cfrac {1}{6} = \cfrac {1}{36} $$

$$ E \left(XY \right)=E \left(X \right) \bullet E\left(Y \right) $$

Example: Expectation of Two Independent Random Variables

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

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

Solution

We know that for two independent random variables:

$$
{E}\left( {X}+ {Y}\right)= {E}\left( {X}\right)+ {E}\left( {Y}\right)
$$

Now,

$$
{E}\left( {X}\right)=\frac{2+0}{2}=1
$$

And

$$
{E}\left( {Y}\right)=\frac{2+0}{2}=1 \\
\Rightarrow\ {E}\left( {X}+ {Y}\right)=1+1=2
$$

For the variance, we know that:

$$
{Var}\left( {X}+ {Y}\right)= {Var}\left(
{X}\right)+ {Var}( {Y})
$$

We have that:

$$
{Var}\left( {X}\right)=\frac{\left(2-0\right)^2}{12}=\frac{1}{3}
$$

And

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

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

$$ \begin{align*}
{f}_\left( {X}+ {Y}\right)\left( {x}\right)&=\begin{cases} \frac { 1 }{ 4 } {x},& 0\le {x}\le 2 \\ \left( \frac { 1 }{ 4 } \right) \left( 4- {x} \right) ,& 2\le {x}\le 4 \\ 0 & \text{otherwise} \\ \end{cases} \\
{E}\left( {E}+ {Y}\right)&=\int_{0}^{2}{\left(\frac{1}{4}\right) {x}. {x}\ {dx}\ +\int_{2}^{4}{\left(\frac{1}{4}\right)\left(4- {x}\right) {x}\ {dx}}} \\
&=\left[\frac{1}{12} {x}^3\right]_{ {x}=0}^{ {x}=2}+\left[\left(\frac{1}{2}\right) {x}^2-\frac{1}{12} {x}^3\right]_{ {x}=2}^{ {x}=4}=\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2\
\end{align*} $$

It is also true that for two independent random variables,

$$
{M}_{ {X}+ {Y}}\left( {t}\right)= {M}_ {X} {t}\times\ {M}_ {Y} {t}
$$

Example: Moment Generating Function of the Two Independent Random Variables

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

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

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

Solution

We know that for two independent random variables:

$$
{M}_{ {X}+ {Y}}\left( {t}\right)= {M}_ {X} {t}\times\ {M}_ {Y} {t}
$$

Thus, in this case:

$$ \begin{align*}
{M}_{ {X}+ {Y}}\left( {t}\right)&= {e}^{0.2\left( {e}^ {t}-1\right)}\times\ {e}^{0.3\left( {e}^ {t}-1\right)} \\
{M}_{ {X}+ {Y}}\left( {t}\right)&= {e}^{(0.2 {e}^ {t}-0.2+0.3 {e}^ {t}-0.3)} \\
&= {e}^{0.5 {e}^ {t}-0.5}= {e}^{0.5\left( {e}^ {t}-1\right)}
\end{align*} $$

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 Poison distribution with parameter \(\lambda=\lambda_ {x}+\lambda_ {y} \).

Example: Sum of Poisson Random Variables

Prove that the sum of two Poisson variables also follows a Poisson distribution.

Solution

The probability generating function (PGF) of a discrete random variable \(x\) is given by:

$$ {G}\left( {t}\right)=\sum_{\left(all\ {x}\right)}{ {P}\left( {X}= {x}\right) {t}^ {x}} $$

Consider \( {X} \sim {Po}(\lambda) \).

Where \( {P}\left( {X}= {x}\right)=\frac{\lambda^ {x} {e}^{-\lambda}}{x!} \).

$$ \begin{align*}
{G}_ {X}\left( {t}\right)&= {E}( {t}^ {x}) \\
&=\sum_{ {x}=0}^{\infty}{ {t}^ {x}\ \frac{\lambda^ {x} {e}^{-\lambda}}{ {x}!}} \\
&= {e}^{-\lambda}\sum_{ {x}=0}^{\infty}\frac{\left( {t}\lambda\right)^ {x}}{ {x}!} \\
&{= {e}}^{-\lambda} {e}^{ {t}\lambda} \\
&{= {e}}^{-\lambda+ {t}\lambda} \\
{ {G}_ {X}( {t})}&= {e}^{-\lambda\left(1+ {t}\right)}
\end{align*} $$

Let \( {Y} \sim {Po}(\mu) \)

$$ {G}_ {Y}\left( {t}\right)= {e}^{-\mu\left(1+ {t}\right)} $$

Thus,

$$ \begin{align*}
{G}_{ {X}+ {Y}}\left( {t}\right)&= {G}_ {X}\left( {t}\right)+ {G}_ {Y}( {t}) \\
&{= {e}}^{-\lambda\left(1+ {t}\right)} {e}^{-\mu\left(1+ {t}\right)} \\
&{= {e}}^{-(\lambda+\mu)\left(1+ {t}\right)}
\end{align*} $$

Hence \( {X}+ {Y}\sim {P}(\lambda+\mu) \).

Example: Sum of the Poisson Random Variables

Let \(X\), \(Y\), and \(Z\) be independent Poisson random variables with \( {E}\left( {X}\right)=1,\ {E}\left( {Y}\right)=2 \text{ and } {E}\left( {Z}\right)=3\).

Find \({P}( {X}+ {Y}+ {Z}\le2) \).

Solution

Fact 1:

The Poisson PMF is given by:

$$ \begin{align*}
& {P}\left( {X}= {x}\right)=\frac{ {e}^{-\lambda}\lambda^ {x}}{ {x}!},\ \ \ {x}=0,1,2,3,\ldots \\
& {E}\left( {X}\right)=\lambda
\end{align*} $$

Fact 2:

If \( {X}\sim \text{ Poisson } (\lambda_1), \)

and \( {Y}\sim \text{ Poisson } \left(\lambda_2\right), {\text{ X and Y iind.}}, \)

then \( {X}+ {Y}\sim \text{ Poisson }(\lambda_1+\lambda_2) \).

So,

$$ \begin{align*}
& {X}+ {Y}+ {Z} \sim Poisson\left(\lambda_1+\lambda_2+\lambda_3\right)=1+2+3=6 \\
& {P}\left( {X}+ {Y}+ {Z}\right)\le2 \\
& {P}\left( {X}+ {Y}+ {Z}=0\right)+ {P}\left( {X}+ {Y}+ {Z}=1\right)+ {P}\left(X+Y+Z=2\right) \\
&=\frac{ {e}^{-6}6^0}{0!}+\frac{ {e}^{-6}6^1}{1!}+\frac{ {e}^{-6}6^2}{2!} \\
&{= {e}}^{-6}(1+6+18) \\
&=25 {e}^{-6}=0.061968804\approx0.062
\end{align*} $$

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 distributed with parameters \(\mu=\mu_ {x}+\mu_ {y}\) and \(\sigma^2=\sigma_ {x}^2+\sigma_ {y}^2\).

Proof:

Suppose that \(X\) is a normal random variable with a given mean and variance so that:

$$ {f}_ {X}\left( {x}\right)=\frac{1}{\sqrt{(2\pi\sigma_ {x})}} {e}^\frac{{-\left( {x}-\mu_ {x}\right)}^2}{2\sigma_ {x}^2} $$

Similarly, \(Y\) is normal with a given mean and variance so that:

$$ {f}_ {Y}\left( {y}\right)=\frac{1}{\sqrt{(2\pi\sigma_ {y})}} {e}^\frac{{-\left( {y}-\mu_ {y}\right)}^2}{2\sigma_ {y}^2} $$

Assume that \(X\) and \(Y\) are independent. We wish to find the sum of the two normal random variables by deriving the PDF of \(Z\):

$$ {Z}= {X}+ {Y} $$

We apply the convolution formulae:

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:

$$ {f}_ {X}\left( {x}\right)=\frac{1}{\sqrt{(2\pi\sigma_ {x})}} {e}^\frac{{-\left( {x}-\mu_ {x}\right)}^2}{2\sigma_ {x}^2}, {f}_ {Y}\left( {y}\right)=\frac{1}{\sqrt{(2\pi\sigma_ {y})}} {e}^\frac{{-\left( {y}-\mu_ {y}\right)}^2}{2\sigma_ {y}^2} $$

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 \( {f}_ {Y}( {Z}- {X}) \).

Therefore,

$$ \begin{align*}
{f}_ {z}( {z}) & =\int _{ -\infty }^{ \infty }{ \cfrac { 1 }{ \sqrt { \left( 2\pi { \sigma }_{ {x} } \right) } } exp } \left\{ -\cfrac { { \left( {x}-{ \mu }_{ {x} } \right) }^{ 2 } }{ 2{ \sigma }_{ {x} }^{ 2 } } \right\} \cfrac { 1 }{ \sqrt { \left( 2\pi { \sigma }_{ {y} } \right) } } exp\left\{ -\cfrac { { \left( {z}- {x}-{ \mu }_{ {y} } \right) }^{ 2 } }{ 2{ \sigma }_{ {y} }^{ 2 } } \right\} {dx} \\
{f}_ {z}( {z}) & =\cfrac { 1 }{ \sqrt { \left( 2\pi \left( { \sigma }_{ {x} }^{ 2 } \right) \right) } } {exp}\left\{ -\cfrac { { \left( {z}-{ \mu }_{ {x} }-{ \mu }_{ {y} } \right) }^{ 2 } }{ 2\left( { \sigma }_{ {x} }^{ 2 }+{ \sigma }_{ {y} }^{ 2 } \right) } \right\}
\end{align*} $$

Note:

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 \(X+Y+W\).

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.

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.

Learning Outcome

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

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