# Show that P(x = x) = \frac{((\lambda_1 + \lambda_2)^x}{x!} e^{-(\lambda_1 + \lambda_2)} for all x...

## Question:

Show that {eq}P(x = x) = \frac{((\lambda_1 + \lambda_2)^x}{x!} e^{-(\lambda_1 + \lambda_2)} {/eq} for all x = 0, 1, 2 ....

Might want to use Binomial Theorem

{eq}(a + b)^n = \sum^n_{k=0} \binom{n}{k}a^kb^{n-k} {/eq}

## Binomial Theorem:

Binomial Theorem is a part of algebraic expansion between the constraints to the power given. In the binomial theorem, the given algebraic expression is presented with the power of expansion and to be arranged in the binomial form of constraints.

## Answer and Explanation:

Given Data

{eq}z = {x_1} + {x_2} {/eq}

Evaluation

{eq}\begin{align*} P\left( {z = x + y} \right) = pz\left( z \right) = \sum\limits_{x = 0}^\theta {px\left( x \right)py\left( {z - x} \right)} = \sum\limits_{x = 0}^\theta {\frac{{{e^{ - {\lambda _1}}}\lambda _1^x}}{{x!}}} \times \frac{{{e^{ - {\lambda _2}}}\lambda _2^{\theta - x}}}{{\left( {z - x} \right)!}}\\ = {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right)}}\sum\limits_{x = 0}^\theta {\frac{{\lambda _2^{\theta - x}\lambda _1^x}}{{\left( {z - x} \right)!x!}} = {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right)}}\sum\limits_{x = 0}^\theta {\frac{{\lambda _2^{\theta - x}\lambda _1^x}}{{\left( {z - x} \right)!x!}} \times \frac{{z!}}{{z!}} = {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right)}}\sum\limits_{x = 0}^\theta {\left( {\frac{z}{x}} \right)} \left( {\frac{{\lambda _2^{\theta - x}\lambda _1^x}}{{z!}}} \right)} } \end{align*} {/eq}

Therefore by binomial theorem:

{eq}{\left( {a + b} \right)^m} = \sum\limits_{n = 0}^m {\left( {\frac{m}{n}} \right){a^n}{b^{m - n}}} {/eq}

So, we have:

{eq}P\left( {z = x + y} \right) = \frac{{{e^{ - \left( {{\lambda _1} + {\lambda _2}} \right)}}}}{{z!}}{\left( {{\lambda _1} + {\lambda _2}} \right)^\theta } {/eq}

Therefore the poison distribution parameter for binomial theorem is:

{eq}{\lambda _1} + {\lambda _2} {/eq}