For example, a variable of Normal distribution, $T$, with mean $\mu$ and variance $\sigma^2$ can be standardized into $S$ like this: $$ S=\frac{T-\mu}{\sigma}\;\Longrightarrow\;F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right) $$

My question is, for the Poisson distribution with probability function $$ f(k;\lambda)=\Pr(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}. $$

Is there a way to standardize the $X$ if I define the standard Poisson Distribution as the distribution that $\lambda=1$?


Yes, there is a standard Poisson, the one with parameter $1$. Recall that if $X$ counts the number of "accidents" in a unit time interval, then under suitable conditions $X$ has Poisson distribution with parameter the mean number of accidents per unit time.

If that parameter is $\lambda$, then the number of accidents in a time interval $t$ is Poisson with parameter $\lambda t$.

We adjust the interval of time over which we count, until we get a time that gives mean count $1$. Of course $t=\frac{1}{\lambda}$ is that time interval.

Like in the standardization of the normal, a change of scale is involved. But in the case of the Poisson it is not the random variable that is being scaled.

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    $\begingroup$ But if $X\sim N(\mu,\sigma^2)$ then $(X-\mu)/\sigma\sim N(0,1)$. There's nothing similar you can do with the Poisson distribution, i.e. no one-to-one function of a Poisson-distributed variable that has this "standard" Poisson distribution, unless the Poisson distribution you started with is already standard. $\endgroup$ – Michael Hardy Jun 28 '13 at 2:44

Let X have a Poisson distribution with parameter λ. Then it can be shown that the random variable Y = (X - λ)/√(λ) will converge to a standard normal distribution as λ goes to infinity. So if you ever have a possion distribution with a relatively large λ, you can standardize it into Y, which will (approximately) have a standard normal distribution.


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