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

AI Thread Summary
The discussion focuses on the approximate distributions of the sample means $\bar{X}$ and $\bar{Y}$ derived from independent Poisson and Normal random variables, respectively. It is established that $\bar{X}$ can be approximated by a normal distribution, specifically $\bar{X} \sim N(2, \frac{2}{35})$, while $\bar{Y} \sim N(1, \frac{2}{15})$. The sum of these two distributions, $\bar{X} + \bar{Y}$, is also approximated as a normal distribution with parameters $N(3, \frac{4}{21})$. Participants also discuss the application of the Central Limit Theorem and validate their findings through simulation in R, confirming the theoretical values for means and variances. The conversation highlights the importance of understanding the underlying distributions and the implications of approximating Poisson distributions with normal ones.
lerem456
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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.
 
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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 $$\frac{2}{35}=0.05714286$$
 
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 $$\frac{2}{35}=0.05714286$$

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.
 
  • #10
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?
 
  • #11
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?

For part (d) I made an addition error.

$\bar{X} \sim N(70, 2)$
$\bar{Y} \sim N(15, 2)$
$\bar{X} + \bar{Y} = \sim N(70+15, 2+2) = \sim N(90, 4)$
 
  • #12
lerem456 said:
For part (d) I made an addition error.

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

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 $$\sum_{i=1}^{n}X_i \sim \text{Poisson }(n\lambda)$$
 
  • #13
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 $$\sum_{i=1}^{n}X_i \sim \text{Poisson }(n\lambda)$$

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)
$\bar{X} + \bar{Y} \sim N(2 + 2, \frac{2}{35} + \frac{2}{15}) = \sim N(4, \frac{4}{21})$
 
  • #14
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 $$\frac{4}{21}=0.1904762$$)
 
  • #15
Thank you Jameson and chisigma for your help. The material makes more sense now.
 
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