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This problem based on Central Limit Theorem gives a detailed solution to ISI M.Stat 2018 PSB Problem 7, with a tinge of simulation and code.

Suppose is a random sample from a bivariate normal distribution with

and unknown for all Define a) Is an unbiased estimator of Justify your answer.

(b) For large obtain an approximate level two-sided confi-

dence interval for where .

- Probability Theory (Expectation, Variance, Covariance, Correlation Coefficient)
- Unbiased Estimator
- Moments of Univariate Normal Distribution
- Bivariate Normal Distribution and a Different Definition
- Central Limit Theorem

**(a)**

Just compute the ).

= = .

.

So, is unbiased for .

**(b) **

Observe that and are independent sample and therefore iid.

So, and are also iid.

Hence, computing the limiting distribution of , flashes in our minds, the Central Limit Theorem. So, let's dig into it. But, for that we need the following:

- =

So, how to calculate the . For that

Two random variables and are said to be jointly normal if they can be expressed in the form , where and are independent standard normal random variables.

Alternate Definition of Bivariate Normal

Why do we need this? Because, and are not independent and they have a correlation coefficient between them.

Assume, ~ .

**Exercise**: Using the above result, prove that can be written as , where ~ N(0,1) and is independent of .

.

**Exercise**: Justify the above steps, using the independence of and .

We used the fact that if ~ N(0,1). Instead of computing the whole we will use the fact that and if ~ .

**Exercise**: Prove that if ~ N(0,1) using the above hint that ~ .

The final result, we got is the following:

.

.

Now use Central Limit Theorem.

Therefore, .

So, . Now, you have to square it to get a confidence interval for .

But, we can use variance stablizing transformation (pivotal method).

Observe that , which is an increasing and hence bijective function.

. Calculate this constanc

Now, try to find a confidence interval for based on this. Then take the inverse of to get a confidence interval for .

```
N <- 2000 # Number of random samples
# Target parameters for univariate normal distributions
v = NULL
rho <- 0.5
mu1 <- 0; s1 <- 1
mu2 <- 0; s2 <- 1
mu <- c(mu1,mu2) # Mean
sigma <- matrix(c(s1^2, s1*s2*rho, s1*s2*rho, s2^2),
2) # Covariance matrix
library(MASS)
for (i in 1:1000) {
bvn1 <- mvrnorm(N, mu = mu, Sigma = sigma ) # from MASS package
W = bvn1[,1]*bvn1[,2]
Wbar = mean(W)
v = c(v, Wbar)
}
hist(v, freq = F)
sigma2 = sqrt(1 + 2*rho^2)/sqrt(N)
x = seq(0.4, 0.6, 0.00001)
curve(dnorm(x, rho, sigma2), from = 0, col = "red", add = TRUE)
```

This problem was a bit more mathematical and technical, but still, I hope that the simulation along with the proofs gave you a good reading experience. Stay Tuned!

This problem based on Central Limit Theorem gives a detailed solution to ISI M.Stat 2018 PSB Problem 7, with a tinge of simulation and code.

Suppose is a random sample from a bivariate normal distribution with

and unknown for all Define a) Is an unbiased estimator of Justify your answer.

(b) For large obtain an approximate level two-sided confi-

dence interval for where .

- Probability Theory (Expectation, Variance, Covariance, Correlation Coefficient)
- Unbiased Estimator
- Moments of Univariate Normal Distribution
- Bivariate Normal Distribution and a Different Definition
- Central Limit Theorem

**(a)**

Just compute the ).

= = .

.

So, is unbiased for .

**(b) **

Observe that and are independent sample and therefore iid.

So, and are also iid.

Hence, computing the limiting distribution of , flashes in our minds, the Central Limit Theorem. So, let's dig into it. But, for that we need the following:

- =

So, how to calculate the . For that

Two random variables and are said to be jointly normal if they can be expressed in the form , where and are independent standard normal random variables.

Alternate Definition of Bivariate Normal

Why do we need this? Because, and are not independent and they have a correlation coefficient between them.

Assume, ~ .

**Exercise**: Using the above result, prove that can be written as , where ~ N(0,1) and is independent of .

.

**Exercise**: Justify the above steps, using the independence of and .

We used the fact that if ~ N(0,1). Instead of computing the whole we will use the fact that and if ~ .

**Exercise**: Prove that if ~ N(0,1) using the above hint that ~ .

The final result, we got is the following:

.

.

Now use Central Limit Theorem.

Therefore, .

So, . Now, you have to square it to get a confidence interval for .

But, we can use variance stablizing transformation (pivotal method).

Observe that , which is an increasing and hence bijective function.

. Calculate this constanc

Now, try to find a confidence interval for based on this. Then take the inverse of to get a confidence interval for .

```
N <- 2000 # Number of random samples
# Target parameters for univariate normal distributions
v = NULL
rho <- 0.5
mu1 <- 0; s1 <- 1
mu2 <- 0; s2 <- 1
mu <- c(mu1,mu2) # Mean
sigma <- matrix(c(s1^2, s1*s2*rho, s1*s2*rho, s2^2),
2) # Covariance matrix
library(MASS)
for (i in 1:1000) {
bvn1 <- mvrnorm(N, mu = mu, Sigma = sigma ) # from MASS package
W = bvn1[,1]*bvn1[,2]
Wbar = mean(W)
v = c(v, Wbar)
}
hist(v, freq = F)
sigma2 = sqrt(1 + 2*rho^2)/sqrt(N)
x = seq(0.4, 0.6, 0.00001)
curve(dnorm(x, rho, sigma2), from = 0, col = "red", add = TRUE)
```

This problem was a bit more mathematical and technical, but still, I hope that the simulation along with the proofs gave you a good reading experience. Stay Tuned!

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The confidence interval contains the unknown parameter i.e. correlation coefficient(row). How?

First of all, this is a large sample confidence interval. With respect to the large sample, the expectation of Wn when n large goes to rho. Hence, we are seeing the confidence interval around the mean. Just, expand that out. You will get the expression. See I have added a new portion. Thanks for your doubt. Stay tuned.

Why is the variance of X1Y1 = E(X1Y1)^2?

E(X1Y1) is rho and not zero

Variance of X1Y1 should be 1+rho^2.