Mixing
Compute the Mean Squared Error of $𝑇$
$begingroup$
The famous biologist Henk de Rijn has come up with a skill test for chimpanzees.
We assume that each chimpanzee has a probability $𝑝$ to pass the test, independent from previous attempts and from other monkeys.
Let $𝑋_𝑖$ be the number of attempts that chimpanzee $𝑖$ needs to pass. Furthermore, $𝑇 = overline{X}_𝑛$ is an estimator for $1/𝑝$.
Compute the mean squared error of $𝑇$ .
So let $T$ be an estimator for a parameter $1∕𝑝$. The $MSE$ of $T$ is $MSE(T)=operatorname{Var}(T) + (E[T] − 1∕𝑝)^2$, where$(E[T] − 1∕𝑝)^2$ is the bias.
In this case the $E[T]=E[overline{X}_𝑛]=mu$ and $operatorname{Var}(T)=operatorname{Var}(overline{X}_𝑛)=dfrac{sigma^2}{n}$.
Now my question is: which distribution should I use?
I think that I think that is a Binomial but it also make sense to use a Geometric Distribution, so then the estimator $T$ will be unbiased and the $displaystyleoperatorname{MSE}(T)=operatorname{Var}(T)=frac{1-p}{p^2}*frac{1}{n}$, is it right?
statistics
edited Jan 23 at 21:33
Adrian Keister
5,27371933
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
$endgroup$
add a comment |
$begingroup$
The famous biologist Henk de Rijn has come up with a skill test for chimpanzees.
We assume that each chimpanzee has a probability $𝑝$ to pass the test, independent from previous attempts and from other monkeys.
Let $𝑋_𝑖$ be the number of attempts that chimpanzee $𝑖$ needs to pass. Furthermore, $𝑇 = overline{X}_𝑛$ is an estimator for $1/𝑝$.
Compute the mean squared error of $𝑇$ .
So let $T$ be an estimator for a parameter $1∕𝑝$. The $MSE$ of $T$ is $MSE(T)=operatorname{Var}(T) + (E[T] − 1∕𝑝)^2$, where$(E[T] − 1∕𝑝)^2$ is the bias.
In this case the $E[T]=E[overline{X}_𝑛]=mu$ and $operatorname{Var}(T)=operatorname{Var}(overline{X}_𝑛)=dfrac{sigma^2}{n}$.
Now my question is: which distribution should I use?
I think that I think that is a Binomial but it also make sense to use a Geometric Distribution, so then the estimator $T$ will be unbiased and the $displaystyleoperatorname{MSE}(T)=operatorname{Var}(T)=frac{1-p}{p^2}*frac{1}{n}$, is it right?
statistics
edited Jan 23 at 21:33
Adrian Keister
5,27371933
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
$endgroup$
1
$begingroup$
Here's what I see.
$endgroup$
– Shaun
Jan 23 at 22:29
$begingroup$
Strange, I see all the value inside the dollars
$endgroup$
– Luke Marci
Jan 24 at 8:19
add a comment |
$begingroup$
The famous biologist Henk de Rijn has come up with a skill test for chimpanzees.
We assume that each chimpanzee has a probability $𝑝$ to pass the test, independent from previous attempts and from other monkeys.
Let $𝑋_𝑖$ be the number of attempts that chimpanzee $𝑖$ needs to pass. Furthermore, $𝑇 = overline{X}_𝑛$ is an estimator for $1/𝑝$.
Compute the mean squared error of $𝑇$ .
So let $T$ be an estimator for a parameter $1∕𝑝$. The $MSE$ of $T$ is $MSE(T)=operatorname{Var}(T) + (E[T] − 1∕𝑝)^2$, where$(E[T] − 1∕𝑝)^2$ is the bias.
In this case the $E[T]=E[overline{X}_𝑛]=mu$ and $operatorname{Var}(T)=operatorname{Var}(overline{X}_𝑛)=dfrac{sigma^2}{n}$.
Now my question is: which distribution should I use?
I think that I think that is a Binomial but it also make sense to use a Geometric Distribution, so then the estimator $T$ will be unbiased and the $displaystyleoperatorname{MSE}(T)=operatorname{Var}(T)=frac{1-p}{p^2}*frac{1}{n}$, is it right?
statistics
edited Jan 23 at 21:33
Adrian Keister
5,27371933
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
$endgroup$
The famous biologist Henk de Rijn has come up with a skill test for chimpanzees.
We assume that each chimpanzee has a probability $𝑝$ to pass the test, independent from previous attempts and from other monkeys.
Let $𝑋_𝑖$ be the number of attempts that chimpanzee $𝑖$ needs to pass. Furthermore, $𝑇 = overline{X}_𝑛$ is an estimator for $1/𝑝$.
Compute the mean squared error of $𝑇$ .
So let $T$ be an estimator for a parameter $1∕𝑝$. The $MSE$ of $T$ is $MSE(T)=operatorname{Var}(T) + (E[T] − 1∕𝑝)^2$, where$(E[T] − 1∕𝑝)^2$ is the bias.
In this case the $E[T]=E[overline{X}_𝑛]=mu$ and $operatorname{Var}(T)=operatorname{Var}(overline{X}_𝑛)=dfrac{sigma^2}{n}$.
Now my question is: which distribution should I use?
I think that I think that is a Binomial but it also make sense to use a Geometric Distribution, so then the estimator $T$ will be unbiased and the $displaystyleoperatorname{MSE}(T)=operatorname{Var}(T)=frac{1-p}{p^2}*frac{1}{n}$, is it right?
statistics
statistics
edited Jan 23 at 21:33
Adrian Keister
5,27371933
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
edited Jan 23 at 21:33
Adrian Keister
5,27371933
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
edited Jan 23 at 21:33
Adrian Keister
5,27371933
edited Jan 23 at 21:33
Adrian Keister
5,27371933
edited Jan 23 at 21:33
Adrian Keister
5,27371933
5,27371933
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
asked Jan 23 at 21:27
Luke MarciLuke Marci
856
856
1
$begingroup$
Here's what I see.
$endgroup$
– Shaun
Jan 23 at 22:29
$begingroup$
Strange, I see all the value inside the dollars
$endgroup$
– Luke Marci
Jan 24 at 8:19
add a comment |
1
$begingroup$
Here's what I see.
$endgroup$
– Shaun
Jan 23 at 22:29
$begingroup$
Strange, I see all the value inside the dollars
$endgroup$
– Luke Marci
Jan 24 at 8:19
1
1
$begingroup$
Here's what I see.
$endgroup$
– Shaun
Jan 23 at 22:29
$begingroup$
Here's what I see.
$endgroup$
– Shaun
Jan 23 at 22:29
$begingroup$
Strange, I see all the value inside the dollars
$endgroup$
– Luke Marci
Jan 24 at 8:19
$begingroup$
Strange, I see all the value inside the dollars
$endgroup$
– Luke Marci
Jan 24 at 8:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Each $X_i$ follows a Geometric distribution.
ie $X_i ~ Geom(p)$, and $T=frac{X_1+...+X_n}{n}$, and the $X_i$s are independent and identically distributed, ie $E(T)=frac{1}{p}$.
Therefore, $MSE(T)$ is just $Var(T)$, and since the variance of a sum is the sum of the variances when dealing with independent random variables, then $Var(T)=Var(frac{X_1+...+X_n}{n})=frac{1}{n^2}$$(Var(X_1)+...+Var(X_n))=frac{1}{n}frac{p}{(1-p)^2}$.
So yes, it is correct.
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
$endgroup$
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
add a comment |
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1 Answer
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oldest
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$begingroup$
Each $X_i$ follows a Geometric distribution.
ie $X_i ~ Geom(p)$, and $T=frac{X_1+...+X_n}{n}$, and the $X_i$s are independent and identically distributed, ie $E(T)=frac{1}{p}$.
Therefore, $MSE(T)$ is just $Var(T)$, and since the variance of a sum is the sum of the variances when dealing with independent random variables, then $Var(T)=Var(frac{X_1+...+X_n}{n})=frac{1}{n^2}$$(Var(X_1)+...+Var(X_n))=frac{1}{n}frac{p}{(1-p)^2}$.
So yes, it is correct.
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
$endgroup$
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
add a comment |
$begingroup$
Each $X_i$ follows a Geometric distribution.
ie $X_i ~ Geom(p)$, and $T=frac{X_1+...+X_n}{n}$, and the $X_i$s are independent and identically distributed, ie $E(T)=frac{1}{p}$.
Therefore, $MSE(T)$ is just $Var(T)$, and since the variance of a sum is the sum of the variances when dealing with independent random variables, then $Var(T)=Var(frac{X_1+...+X_n}{n})=frac{1}{n^2}$$(Var(X_1)+...+Var(X_n))=frac{1}{n}frac{p}{(1-p)^2}$.
So yes, it is correct.
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
$endgroup$
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
add a comment |
$begingroup$
Each $X_i$ follows a Geometric distribution.
ie $X_i ~ Geom(p)$, and $T=frac{X_1+...+X_n}{n}$, and the $X_i$s are independent and identically distributed, ie $E(T)=frac{1}{p}$.
Therefore, $MSE(T)$ is just $Var(T)$, and since the variance of a sum is the sum of the variances when dealing with independent random variables, then $Var(T)=Var(frac{X_1+...+X_n}{n})=frac{1}{n^2}$$(Var(X_1)+...+Var(X_n))=frac{1}{n}frac{p}{(1-p)^2}$.
So yes, it is correct.
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
$endgroup$
Each $X_i$ follows a Geometric distribution.
ie $X_i ~ Geom(p)$, and $T=frac{X_1+...+X_n}{n}$, and the $X_i$s are independent and identically distributed, ie $E(T)=frac{1}{p}$.
Therefore, $MSE(T)$ is just $Var(T)$, and since the variance of a sum is the sum of the variances when dealing with independent random variables, then $Var(T)=Var(frac{X_1+...+X_n}{n})=frac{1}{n^2}$$(Var(X_1)+...+Var(X_n))=frac{1}{n}frac{p}{(1-p)^2}$.
So yes, it is correct.
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
answered Jan 23 at 21:56
AlexandrosAlexandros
9401412
9401412
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
add a comment |
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
$begingroup$
The Variance for a geometric distribution is not $frac{1-p}{p^2}$?
$endgroup$
– Luke Marci
Jan 24 at 8:30
add a comment |
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1
$begingroup$
Here's what I see.
$endgroup$
– Shaun
Jan 23 at 22:29
$begingroup$
Strange, I see all the value inside the dollars
$endgroup$
– Luke Marci
Jan 24 at 8:19