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
.
(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
Alternate Definition of Bivariate Normaland
are said to be jointly normal if they can be expressed in the form
, where
and
are independent standard normal random variables.
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
.
(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
Alternate Definition of Bivariate Normaland
are said to be jointly normal if they can be expressed in the form
, where
and
are independent standard normal random variables.
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!
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.