Q. 8.14 - A First Course in Probability

Question

 Let $X_1,X_2,\dots$be a sequence of independent and identically distributed random variables with distribution$F$, having a finite mean and variance. Whereas the central limit theorem states that the distribution of$\sum _{i=1}^n{X}_i$ approaches a normal distribution as $n$goes to infinity, it gives us no information about how large$n$ need to be before the normal becomes a good approximation. Whereas in most applications, the approximation yields good results whenever$n\ge 20$, and oftentimes for much smaller values of$n$, how large a value of $n$is needed depends on the distribution of$X_i$. Give an example of distribution $F$such that the distribution$\sum _{i=1}^{100}{X}_i$ is not close to a normal distribution.

Hint: Think Poisson.

Step-by-Step Solution

Answer

Therefore,

 we can conclude that the normal approximation$\sum _{i=1}^{100}{X}_i$ is not good.

Step 1 Given Information.

$X1, X2, ...$be a sequence of independent and identically distributed random variables with the distribution$F$.

Step 2 Explanation.

Using the central limit theorem, we have that

$\sum _{i=1}^{100}{X}_i~AN\left(100\mu ,{\sigma }^2\right)$

where$\mu =E\left(X_1\right)$ and${\sigma }^2=Var\left(X_1\right)$. Now suppose that$X_i~Pois\left(\lambda \right)$. Therefore$\mu ={\sigma }^2=\lambda$. We know that the sum of independent Poisson distributions is Poisson distribution. Therefore

$\sum _{i=1}^{100}{X}_i~Pois\left(100\lambda \right)$

which, implies that$E\left(\sum _{i=1}^{100}{X}_i\right)=Var\left(\sum _{i=1}^{100}{X}_i\right)=100\lambda$. On the other hand,$CLT$ gives us an approximation

$\sum _{i=1}^{100}{X}_i~AN\left(100\mu ,{\sigma }^2\right)=AN(100\lambda ,\lambda )$

so the approximations for mean and variance are $E\left(\sum _{i=1}^{100}{X}_i\right)\approx 100\lambda$ and$Var\left(\sum _{i=1}^{100}{X}_i\right)=\lambda$. Even though the approximation for the mean is exact, the approximation for the variance is significantly wrong since$100\lambda \not\approx \lambda$. Therefore, we can conclude that the normal approximation $\sum _{i=1}^{100}{X}_i$is not good.