Let X be a random variable which is Poisson distributed with parameter λ. Show that E(X) = λ
We are given the following hint:
\[ e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+.... \]
Our hint is the first few terms of the series for exponential function https://en.wikipedia.org/wiki/Exponential_function
We now that the expectation of a random variable is found by
\[ E(x)=\sum { xf(x) } \]
We are told our random variable is poisson distributed with parameter lambda
\[ f(x)=\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } \] Plug in the expression for f(x) into the formula for expected value
\[ E(x)=\sum { x\frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } } \]
What happens if we expand a few terms of this series with index x starting at 0?
\[ E(x)=0\frac { { e }^{ -\lambda }{ \lambda }^{ 0 } }{ 0! }+1\frac { { e }^{ -\lambda }{ \lambda }^{ 1 } }{ 1! }+2\frac { { e }^{ -\lambda }{ \lambda }^{ 2 } }{ 2! }+3\frac { { e }^{ -\lambda }{ \lambda }^{ 3 } }{ 3! }... \]
Notice that every term has something in common. We can remove the eponential term and the lambda term. Note: the first term in our series is zero
Factor out a lambda and the exponent from each term
\[ E(x)=e ^{ -\lambda }\lambda(1+\lambda+\frac{\lambda^{2}}{2!}+\frac{\lambda^{3}}{3!}+...) \]
Recall our hint
\[ e^{\lambda}=1+\lambda+\frac{\lambda^{2}}{2!}+\frac{\lambda^{3}}{3!}+... \]
The terms in the () are nothing other than the infinite series for the expinential function using lambda instead of x. We can substitue accordingly
\[ E(x)=e ^{ -\lambda }\lambda(e^{\lambda})\\ =e^{-\lambda+\lambda}\lambda\\ =\lambda \]
Hence proved