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This problem based on the calculation of Mean Square Error gives a detailed solution to ISI M.Stat 2019 PSB Problem 5, with a tinge of simulation and code.

Suppose are independent random variables such that

where are all distinct and unknown. Consider and another random variable which is distributed as Binomial where Between and which is a better estimator of in terms of their respective mean squared errors?

- Bernoulli and Binomial Distribution
- Basic Estimation Theory (Unbiasedness and Mean Square Error)
- Cauchy - Schwartz Inequality

.

~ Binomial

.

If is unbiased for , then MSE(.

Observe that .

This results in the fact that .

Therefore, is a better estimate thatn w.r.t Mean Square Error.

Let's verify this as usual by simulation.

```
library(statip)
library(Metrics)
N = 10
p = runif(10, 0, 1)
X = rep(0,N)
vX = NULL
vY = NULL
for (j in 1:1000)
{
for (i in 1:N)
{
X[i] = rbern(1,p[i])
}
Z = sum(X) #sum of Xi random variables
Y = rbinom(1,N,mean(p)) #Y random variable
vX = c(vX, Z)
vY = c(vY, Y)
}
k = rep(sum(p), 1000)
mse(k, vX) #MSE of Sum Xi #1.57966
mse(k, vY) #MSE of Y #2.272519
```

Hence, the theory is verified by this simulation. I hope it helps.

This problem based on the calculation of Mean Square Error gives a detailed solution to ISI M.Stat 2019 PSB Problem 5, with a tinge of simulation and code.

Suppose are independent random variables such that

where are all distinct and unknown. Consider and another random variable which is distributed as Binomial where Between and which is a better estimator of in terms of their respective mean squared errors?

- Bernoulli and Binomial Distribution
- Basic Estimation Theory (Unbiasedness and Mean Square Error)
- Cauchy - Schwartz Inequality

.

~ Binomial

.

If is unbiased for , then MSE(.

Observe that .

This results in the fact that .

Therefore, is a better estimate thatn w.r.t Mean Square Error.

Let's verify this as usual by simulation.

```
library(statip)
library(Metrics)
N = 10
p = runif(10, 0, 1)
X = rep(0,N)
vX = NULL
vY = NULL
for (j in 1:1000)
{
for (i in 1:N)
{
X[i] = rbern(1,p[i])
}
Z = sum(X) #sum of Xi random variables
Y = rbinom(1,N,mean(p)) #Y random variable
vX = c(vX, Z)
vY = c(vY, Y)
}
k = rep(sum(p), 1000)
mse(k, vX) #MSE of Sum Xi #1.57966
mse(k, vY) #MSE of Y #2.272519
```

Hence, the theory is verified by this simulation. I hope it helps.

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