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# 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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