Tag: random variables and probability distribution

$P(X>1)$
$=1-[P(X=0)+P(X=1)]$
$=1-[e^{-1.5}+1.5e^{-1.5}]$
$=1-e^{1.5}(1+1.5)$
$=1-e^{1.5}(2.5)$

$P(X=0)=P(X=1) $
P$(0; \mu) = \dfrac { { e }^{-\mu}{\mu}^{0}}{0!} =  P(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!}$
$ { e }^{-\mu}  (1 - \mu )$ = 0 
 $\mu = 1 $
P(X=2) = $ \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \dfrac{{ e }^{-1}}{2} = \dfrac {1}{2e} $

$P(X=k)=P(X=k+1)$
$ P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! } $

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

For Poisson's distribution, $P(X)$ is given by 
$\displaystyle P(X)=\frac { { e }^{ -\lambda  }.{ \lambda  }^{ x } }{ x! } $
Hence $\displaystyle P(X=1) =\displaystyle  \frac { { e }^{ -\lambda }.{ \lambda  }^{ 1 } }{ 1! } $.
Similarly $\displaystyle P(X=2)=\frac { { e }^{ -\lambda }.{ \lambda  }^{ 2 } }{ 2! } $.
Given that, $P(X=1) = 2 P(X=2)$
$\Rightarrow \displaystyle \frac { { e }^{ -\lambda }.{ \lambda  }^{ 1 } }{ 1! } =2.\frac { { e }^{ -\lambda }.{ \lambda  }^{ 2 } }{ 2! } $
Simplifying we get $\lambda = 1$

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$
$P(X=1) = P(X=2)$

$ P(X=1)=P(X=2)$
 $P$$(1; \mu) = \dfrac { { e }^{-\mu}{\mu}^{1}}{1!} =  P(2; \mu) = \dfrac { { e }^{-\mu}{\mu}^{2}}{2!}$
$ { e }^{-\mu} [\mu (2 - \mu )]$ $= 0$

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=0)=P(x=1)$
$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=\dfrac{\lambda^{1}e^{-\lambda}}{1!}$
$\lambda=1$
Hence $P(X=0)=P(X=1)$
$=\dfrac{1}{e}$

mean of P.D. $=np = 8$
And $P(X = 4) = \cfrac{e^{-8}(8)^4}{4!}=\cfrac{512}{3e^8}$
Hence both statement are correct but they are not related to each other.

Poisson's distribution implies
$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=$\lambda$
Hence $2.P(x=0)+P(x=2)=2P(x=1)$
$2\dfrac{\lambda^{0}e^{-\lambda}}{0!}+\dfrac{\lambda^{2}e^{-\lambda}}{2!}=2\dfrac{\lambda^{1}e^{-\lambda}}{1!}$