Question

Let the random variable $Z_n$ have a Poisson distribution with parameter $\mu =n$. Show that the limiting distribution of the random variable $Y_n=(Z_n-n)/\sqrt{n}$ is normal with mean zero and variance $1.$

Short answer

Following the calculations and application of the Central Limit Theorem, it was found that as n tends to infinity, the mean and variance of $Y_n$ approach 0 and 1, respectively. Thus, the limiting distribution of $Y_n$ is a standard normal distribution, i.e., a normal distribution with mean 0 and variance 1.

Step-by-step solution

Calculate the Mean of Yn

First, it's crucial to calculate the mean value of $Y_n$. Due to the properties of the expectation operator, we have $E[Y_n]=E[(Z_n-n)/\sqrt{n}]=E[Z_n]/\sqrt{n}-n/\sqrt{n}$. Since $Z_n$ follows a Poisson distribution with parameter $n$, its expected value $E[Z_n]$ is just $n$. Hence, $E[Y_n]=n/\sqrt{n}-n/\sqrt{n}=0$. As we can see, the mean of $Y_n$ is 0, which aligns with the requirement.

Calculate the Variance of Yn

Next, the variance of $Y_n$ needs to be calculated. Again due to the properties of the variance operator, we have $Var[Y_n]=Var[(Z_n-n)/\sqrt{n}]=Var[Z_n]/n-Var[n]/n=Var[Z_n]/n$. Considering that $Z_n$ is Poisson distributed with parameter $n$, its variance $Var[Z_n]$ is $n$. Hence, $Var[Y_n]=n/n=1$, which also aligns with the requirement.

Apply the Central Limit Theorem

Finally, apply the Central Limit Theorem (CLT) which states that the sum of a large number of independent and identically distributed variables will be approximately normally distributed. Here, $Z_n$ can be considered as the sum of $n$ independent Poisson(1) random variables. Therefore, as $n$ tends to infinity, the central limit theorem implies that the distribution of $Y_n$ converges to a normal distribution. Considering that we have shown both mean and variance of $Y_n$ are as required, we can conclude that the limiting distribution of $Y_n$ is a standard normal distribution (mean = 0, variance = 1).