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

[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.

 

[chisigma]

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

$\chi$ $\sigma$

 

[lerem456]

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 $\bar{X_i} \sim N(\lambda_i, \frac{\lambda_i}{n})=N(2, \frac{2}{35})$

 

[chisigma]

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

$\chi$ $\sigma$

 

[lerem456]

... 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?

 

[Jameson]

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.

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.

mean(i)
var(i)

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

 

[chisigma]

I can't approximate using a normal distribution?

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

Kind regards

$\chi$ $\sigma$

 

[lerem456]

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.

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.

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.

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.

 

[Jameson]

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.

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.

 

[Jameson]

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?

 

[lerem456]

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.

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

 

[Jameson]

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 \(\displaystyle \sum_{i=1}^{n}X_i \sim \text{Poisson }(n\lambda)\)

 

[lerem456]

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

 

[Jameson]

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.

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 \(\displaystyle \frac{4}{21}=0.1904762\))

 

[lerem456]

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