Tag: random variables and probability distribution

In  Poisson distribution such that  $ \alpha = p(X=1)=p(X=2) $ 
        $   p(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $  
       => $ \alpha = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}  $       =>   $ \alpha =  { e }^{-\mu}{\mu}  $
                 =>  $ { e }^{-\mu}{\mu} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2} $
                 =>   $ \mu = 2 $ 
  $   p(4; \mu) = \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} =  \dfrac { { e }^{ -\mu  }{ \mu \times { \mu  }^{ 3 } } }{ 24 }  = \dfrac { \alpha {\times { 2 }^{ 3 } } }{ 24 }  = \dfrac { \alpha  }{ 3 } $  

$E[{ X }^{ 2 }]= \sum _{ k=0 }^{ \infty  }{ k^{ 2 } } \sum _{  }^{  }{  } \dfrac { 1 }{ k! } \lambda ^{ k }e^{ -\lambda  }\ \ = ^{ k }e^{ -\lambda  }\sum _{ 1 }^{ \infty  }{ k\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } \= ^{ k }e^{ -\lambda  }(\sum _{ 1 }^{ \infty  }{ (k-1)\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } ) +\sum _{ 1 }^{ \infty  }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ =^{ k }e^{ -\lambda  }(\sum _{ 2 }^{ \infty  }{ (\lambda \dfrac { 1 }{ (k2)! } \lambda ^{ k-2 } } )+ \sum _{ 1 }^{ \infty  }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\ = ^{ k }e^{ -\lambda  }(\sum _{ i=0 }^{ \infty  }{ (\lambda \dfrac { 1 }{ i! } \lambda ^{ i } } )+ \sum _{ j=0 }^{ \infty  }{ \dfrac { 1 }{ j! } \lambda ^{ j } } )\ = ^{ k }e^{ -\lambda  }(\lambda e^{ \lambda  }+e^{ \lambda  })\  \ = { \lambda  }^{ 2 }+ \lambda \ \ Variance= E[{ X }^{ 2 }]- { E[X] }^{ 2 }\ = { \lambda  }^{ 2 }+ \lambda - { \lambda  }^{ 2 }\ = \lambda \ \ \ $

In  Poisson distribution such that $P(X=1)=P(X=2)$
    $ P(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!}$
          =>  $ P(1; \mu)  =  P(2; \mu) $
         =>  $ \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}  $ 
          => $ 1 =$ $ \dfrac {\mu}{2} $ 
         => $ \mu = 2 $
 In Poisson distribution Variance $(m)$ is equal to mean. 
       Mean = Variance  = $2 $

Sum of the terms in Even places will be 
$=e^{-m}[m+\dfrac{m^{3}}{3!}+\dfrac{m^{5}}{5!}+...]$

$=e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]$ ...(i)
Sum of the terms in Odd places will be 
$=e^{-m}[1+\dfrac{m^{2}}{2!}+\dfrac{m^{4}}{4!}+...]$

$=e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]$ ...(ii)
Taking the ratio of i and ii, we get 
$\dfrac{i}{ii}$

$=\dfrac{e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]}{e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]}$

$=\dfrac{e^{m}-e^{-m}}{e^{m}+e^{-m}}$

$=\dfrac{2sinhm}{2coshm}$

$=tanh(m)$.

If m is the variance of P. D, then  P$(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in even places = [ P$(1; \mu) + P(3; \mu)  + P(5; \mu) + .........$ ]

If m is the variance of P. D, then  P$(x; \mu) = \frac { { e }^{-\mu}{\mu}^{x}}{x!} $
Sum of the terms in odd places = [ P$(0; \mu) + P(2; \mu)  + P(4; \mu) + .........$ ]

Considering all odd places in Poisson's Distribution, we get 
$e^{-\lambda}+\dfrac{e^{-\lambda}.\lambda^{2}}{2!}+\dfrac{e^{-\lambda}.\lambda^{4}}{4!}....$

$=e^{-\lambda}[1+\dfrac{\lambda^{2}}{2!}+\dfrac{\lambda^{4}}{4!}+....]$

$=e^{-\lambda}[\dfrac{e^{-\lambda}+e^{\lambda}}{2}]$

$=e^{-\lambda}(\cosh\lambda)$.
Hence reason is false.

Here,  $P(X=0)=P(X=1)$
$\Rightarrow \displaystyle \frac{e^{-\lambda}\lambda^0}{0!}=\frac{e^{-\lambda}\lambda^1}{1!}\Rightarrow \lambda = 1$
$\displaystyle \therefore P(X = 2) = \frac{e^{-\lambda}1^2}{2!}=\frac{1}{2e}$

Variance  = $E(X^{ 2 }) - E[X]^{ 2 }$
Mean = Variance for P.D.
Let mean = $m$
Therefore
$m = $ $ 6 - m*m $
Hence
$m = 2$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=2)=3P(x=3)$
$3\dfrac{\lambda^{3}e^{-\lambda}}{3!}=\dfrac{\lambda^{2}e^{-\lambda}}{2!}$
$3\lambda=3$
$\lambda=1$
Hence mean=variance=$1$