# What Are the Approximate Distributions of $\bar{X}$ and $\bar{Y}$?

• MHB
• lerem456
In summary, the conversation addresses the distribution of independent Poisson and Normal random variables, and the use of the central limit theorem to approximate these distributions. The approximate distributions of $\bar{X}$ and $\bar{Y}$ are given, along with the approximate distribution of $\bar{X} + \bar{Y}$. The conversation also discusses the use of R code to demonstrate the central limit theorem and the calculation of probabilities for $\bar{X}$ and $\bar{Y}$. There is a question about the necessity of using a normal distribution for part (b) instead of the exact expression.
lerem456
Let $X_1, ..., X_{35}$ be independent Poisson random variables having mean and variance 2.

Let $Y_1, ..., Y_{15}$ be independent Normal random variables having mean 1 and variance 2.

(a.) Specify the (approximate) distributions of $\bar{X}$.
(b.) Find the probability $P(1.8 \leq \bar{X} \leq 2.3)$
(c.) Specify the (approximate) distributions of $\bar{Y}$.
(d.) Specify the (approximate distributions of $\bar{X} + \bar{Y}$.
(e.) Find the probability $P(Y_1 + ... + Y_{15} \leq 20)$.

My attempt,
(a.) Since $n > 30$. I can use the central limit theorem to state $\bar{X} \sim N(2, \frac{2}{35})$

(b.) Apparently this one is incorrect and I'm unsure where my error is.
\begin{equation*}\sigma = \sqrt{\frac{2}{35}} = 0.24\end{equation*}
\begin{equation*}P(-\frac{0.2}{0.24} \leq Z \leq \frac{0.3}{0.24}) = P(-0.84 \leq Z \leq 1.25)\end{equation*}
(c.) $\bar{Y} \sim N(1, \frac{2}{15})$

(d.) $\bar{X} + \bar{Y} ~N(35*2 + 15*1, 35*\frac{2}{35} + 15*\frac{2}{15}) = \sim N(50, 4)$

(e.) $P(Z \leq \frac{5}{\sqrt{30}}) = P(Z \leq 0.91)$Any insight on any part is greatly appreciated. Thanks.

Last edited:
lerem456 said:
Let $X_1, ..., X_{35}$ be independent Poisson random variables having mean and variance 2.

Let $Y_1, ..., Y_{15}$ be independent Normal random variables having mean 1 and variance 2.

(a.) Specify the (approximate) distributions of $\bar{X}$.
(b.) Find the probability $P(1.8 \leq \bar{X} \leq 2.3)$
(c.) Specify the (approximate) distributions of $\bar{Y}$.
(d.) Specify the (approximate distributions of $\bar{X} + \bar{Y}$.
(e.) Find the probability $P(Y_1 + ... + Y_{15} \leq 20)$.

My attempt,
(a.) Since $n > 30$. I can use the central limit theorem to state $\bar{X} \sim N(2, \frac{2}{35})$

(b.) Apparently this one is incorrect and I'm unsure where my error is.
\begin{equation*}\sigma = \sqrt{\frac{2}{35}} = 0.24\end{equation*}
\begin{equation*}P(-0.2/0.24 \leq Z \leq 0.3/0.24) = P(-0.84 \leq Z \leq 1.25)\end{equation*}
(c.) $\bar{Y} \sim N(1, \frac{2}{15})$

(d.) $\bar{X} + \bar{Y} ~N(35*2 + 15*1, 35*\frac{2}{35} + 15*\frac{2}{15}) = \sim N(50, 4)$

(e.) $P(Z \leq \frac{5}{\sqrt{30}}) = P(Z \leq 0.91)$Any insight on any part is greatly appreciated. Thanks.

Wellcome on MHB lerem 456!...

... reagarding the point a), if $X_{i}$ are Poisson RV with mean and variance $\lambda_{i}$, $i = 1,2,..., n$ the the RV $X = \sum_{i} X_{i}$ is also Poisson with mean and variance $\lambda = \sum_{i} \lambda_{i}$...

Kind regards

$\chi$ $\sigma$

chisigma said:
Wellcome on MHB lerem 456!...

... reagarding the point a), if $X_{i}$ are Poisson RV with mean and variance $\lambda_{i}$, $i = 1,2,..., n$ the the RV $X = \sum_{i} X_{i}$ is also Poisson with mean and variance $\lambda = \sum_{i} \lambda_{i}$...

Kind regards

$\chi$ $\sigma$

Thanks chisigma.

So the total distribution would be $\bar{X} \sim N(\lambda=\sum_iX_i, \lambda=\sum_iX_i) = N(70, 70)$ where as each individual distribution would be $\bar{X_i} \sim N(\lambda_i, \frac{\lambda_i}{n})=N(2, \frac{2}{35})$

lerem456 said:
Thanks chisigma.

So the total distribution would be $\bar{X} \sim N(\lambda=\sum_iX_i, \lambda=\sum_iX_i) = N(70, 70)$ where as each individual distribution would be $\bar{X_i} \sim N(\lambda_i, \frac{\lambda_i}{n})=N(2, \frac{2}{35})$

... if a distribution is Poisson... then it isn't a normal distribution (Thinking) ...

Kind regards

$\chi$ $\sigma$

chisigma said:
... if a distribution is Poisson... then it isn't a normal distribution (Thinking) ...

Kind regards

$\chi$ $\sigma$

I can't approximate using a normal distribution?

Yes, I believe you can. The central limit theorem says that the mean of a large number of random variables will converge to a normal distribution, regardless of the underlying distribution.

To show this empirically, I ran some code in R. I am generating 35 Poisson random variables and taking the mean of them. This represents our $\bar{X}$. I repeated this 10,000 times so we have a large enough sample to work with.
Code:
i = rep(0,1000)
for (j in 1:1000){
i[j]=mean(rpois(35,2))
}

Now let's look at the mean and variance of $i$. For a sample of 10,000 it should be pretty close to the theoretical value.
Code:
mean(i)
var(i)

The mean of $i$ is 2.001491 and the variance of $i$ is 0.05669634. For reference $$\displaystyle \frac{2}{35}=0.05714286$$

lerem456 said:
I can't approximate using a normal distribution?

Why approximate when You have an exact expression easier to calculate? (Thinking)...

Kind regards

$\chi$ $\sigma$

Jameson said:
Yes, I believe you can. The central limit theorem says that the mean of a large number of random variables will converge to a normal distribution, regardless of the underlying distribution.

To show this empirically, I ran some code in R. I am generating 35 Poisson random variables and taking the mean of them. This represents our $\bar{X}$. I repeated this 10,000 times so we have a large enough sample to work with.
Code:
i = rep(0,1000)
for (j in 1:1000){
i[j]=mean(rpois(35,2))
}

Now let's look at the mean and variance of $i$. For a sample of 10,000 it should be pretty close to the theoretical value.
Code:
mean(i)
var(i)

The mean of $i$ is 2.001491 and the variance of $i$ is 0.05669634. For reference $$\displaystyle \frac{2}{35}=0.05714286$$

Thanks, that makes sense.
chisigma said:
Why approximate when You have an exact expression easier to calculate? (Thinking)...

Kind regards

$\chi$ $\sigma$

Part (b) deals with finding the probability of $\bar{X}$ and part (d) is addition with normal variables. I thought it would be easier to work with two normal distributions but if it's unnecessary then I will look to change my work.

chisigma is correct that the sum of $X_{1}+X_{2}+...X_{35} \sim \text{Poisson }(35\lambda)$. The question is asking about $\bar{X}$ though, which explains the discrepancy. If I modify my previous code we can verify this.
Code:
i = rep(0,1000)
for (j in 1:1000){
i[j]=sum(rpois(35,2))
}
mean(i)
var(i)

Now we get the mean is 70.179 and variance 70.8438, very close to our theoretical values.

Ok, so I agree with your answers for (b) and (c). I don't know how many decimal places they want the answer to have, so maybe you rounded too early.

For (d) I can't follow your work and I get a very different answer. We have two approximately normal distributions added together. What is the mean and variance of the sum of two independent normal random variables?

Jameson said:
Ok, so I agree with your answers for (b) and (c). I don't know how many decimal places they want the answer to have, so maybe you rounded too early.

For (d) I can't follow your work and I get a very different answer. We have two approximately normal distributions added together. What is the mean and variance of the sum of two independent normal random variables?

$\bar{X} \sim N(70, 2)$
$\bar{Y} \sim N(15, 2)$
$\bar{X} + \bar{Y} = \sim N(70+15, 2+2) = \sim N(90, 4)$

lerem456 said:

$\bar{X} \sim N(70, 2)$
$\bar{Y} \sim N(15, 2)$
$\bar{X} + \bar{Y} = \sim N(70+15, 2+2) = \sim N(90, 4)$

Assuming all of the $X_i$ are Poisson random variables with $\lambda =2$ then $\bar{X} \sim N(2, \frac{2}{35})$ right? Remember that the mean of a large number of iid random variable converges to a normal distribution with parameters $\displaystyle \left( \frac{X_1+X_2+...X_n}{n},\frac{\sigma^2}{n}\right)$.

This isn't being asked in this problem but $$\displaystyle \sum_{i=1}^{n}X_i \sim \text{Poisson }(n\lambda)$$

Jameson said:
Assuming all of the $X_i$ are Poisson random variables with $\lambda =2$ then $\bar{X} \sim N(2, \frac{2}{35})$ right? Remember that the mean of a large number of iid random variable converges to a normal distribution with parameters $\displaystyle \left( \frac{X_1+X_2+...X_n}{n},\frac{\sigma^2}{n}\right)$.

This isn't being asked in this problem but $$\displaystyle \sum_{i=1}^{n}X_i \sim \text{Poisson }(n\lambda)$$

I think I understand now. I've noticed inconsistencies in my work in regard to $\bar{X}$.

$\bar{X}\sim N(2, \frac{2}{35})$ from part (a)
$\bar{Y}\sim N(1, \frac{2}{15})$ from part (c)
$\bar{X} + \bar{Y} \sim N(2 + 2, \frac{2}{35} + \frac{2}{15}) = \sim N(4, \frac{4}{21})$

Very close. The mean of the $\bar{X}+\bar{Y}$ is 2+1=3. Just to show you that this can be backed up through simulation, here is R code that shows this.

Code:
x=rep(0,30000)
y=rep(0,30000)

for (j in 1:30000){
x[j]=mean(rpois(35,2))
}

for (i in 1:30000){
y[i]=mean(rnorm(15,1,sqrt(2)))
}

mean(x+y)
var(x+y)

The mean is 2.995854 (very close to 3) and the variance is 0.1905364 (very close to $$\displaystyle \frac{4}{21}=0.1904762$$)

Thank you Jameson and chisigma for your help. The material makes more sense now.

## 1. What are approximate distributions?

Approximate distributions refer to probability distributions that are used to estimate or approximate the true underlying distribution of a dataset. These distributions are often used when the true distribution is unknown or too complex to model accurately. Examples of approximate distributions include the normal approximation, the Poisson approximation, and the beta approximation.

## 2. How are approximate distributions different from exact distributions?

Exact distributions, also known as theoretical distributions, represent the true underlying distribution of a dataset. These distributions are known and can be described by a mathematical formula. On the other hand, approximate distributions are used to estimate the true distribution and are often simpler and easier to work with compared to exact distributions. Approximate distributions are also based on assumptions and may not always accurately reflect the true distribution.

## 3. What are the advantages of using approximate distributions?

One advantage of using approximate distributions is that they can simplify complex problems and make them easier to solve. Additionally, they can provide a good approximation of the true distribution, especially when the underlying distribution is unknown or difficult to model. Approximate distributions can also be used to make predictions and conduct statistical analyses, even when the true distribution is unknown.

## 4. How are approximate distributions used in statistical inference?

Approximate distributions are commonly used in statistical inference to make inferences about a population based on a sample. In this context, they can be used to approximate the sampling distribution of a statistic and make inferences about the population parameter. Approximate distributions are also used in hypothesis testing and confidence interval estimation to make decisions about the population based on the sample data.

## 5. What are some examples of real-world applications of approximate distributions?

Approximate distributions are used in various fields, including finance, economics, biology, and engineering. For example, the normal distribution is often used to approximate stock price movements, and the Poisson distribution is commonly used to model the number of customers arriving at a bank. In biology, the binomial distribution is used to approximate the number of successful outcomes in a series of trials, and in engineering, the exponential distribution is used to model the time between equipment failures.

Replies
1
Views
10K
Replies
1
Views
985
Replies
6
Views
1K
Replies
4
Views
2K
Replies
6
Views
1K
Replies
7
Views
1K
Replies
5
Views
2K
Replies
5
Views
1K
Replies
9
Views
1K
Replies
0
Views
941