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Mean squared error calculation [closed]
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If $ X_1,...,X_n$ ~ $N(mu, sigma^2)$ where $mu$ is known and $sigma^2$ is unknown, calculate the MSE of $V^2$
$V^2 = frac1n sum_X_i^n Var(X_i) =sigma^2$
Therefore:
$MSE(V^2) = Var(V^2) = frac1n^2nVar[(X_1-mu)^2]=frac1nVar[sigma^2(fracX_1-musigma)^2]=frac1nsigma^4Var[(fracX_1-musigma)^2]=frac2sigma^4n$
However, I do not understand some of the steps:
- Where does the $X_1$ suddenly come from (instead of$ X_i$)?
- And then in the next step, I am aware it has somehing to do with the fact that $fracX-musigma$ ~ $chi^2_1$ But i cannot connect the dots .
Could someone break these down for me ? I do not have an mathematical background, therefore stating the obvious is very welcome.
statistics convergence parameter-estimation mean-square-error
asked Jul 15 at 9:43
Danka
178
closed as unclear what you're asking by StubbornAtom, Mostafa Ayaz, José Carlos Santos, user223391, Parcly Taxel Jul 16 at 4:26
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |Â
up vote
1
down vote
favorite
If $ X_1,...,X_n$ ~ $N(mu, sigma^2)$ where $mu$ is known and $sigma^2$ is unknown, calculate the MSE of $V^2$
$V^2 = frac1n sum_X_i^n Var(X_i) =sigma^2$
Therefore:
$MSE(V^2) = Var(V^2) = frac1n^2nVar[(X_1-mu)^2]=frac1nVar[sigma^2(fracX_1-musigma)^2]=frac1nsigma^4Var[(fracX_1-musigma)^2]=frac2sigma^4n$
However, I do not understand some of the steps:
- Where does the $X_1$ suddenly come from (instead of$ X_i$)?
- And then in the next step, I am aware it has somehing to do with the fact that $fracX-musigma$ ~ $chi^2_1$ But i cannot connect the dots .
Could someone break these down for me ? I do not have an mathematical background, therefore stating the obvious is very welcome.
statistics convergence parameter-estimation mean-square-error
asked Jul 15 at 9:43
Danka
178
closed as unclear what you're asking by StubbornAtom, Mostafa Ayaz, José Carlos Santos, user223391, Parcly Taxel Jul 16 at 4:26
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
If $X_i sim N(mu, sigma^2)$ then what is the difference between $Var(X_i)$ and $sigma^2$? Are the $X_i$ independent? Also, what is the meaning of $sum_X_i^n$? Finally, it looks like you are defining $V^2$ as the constant $sigma^2$ so there is no estimation going on and the variance of $V^2$ is 0. I suspect you are incorrectly interpreting a problem you are given, I would expect $V^2$ to be some estimate formed from the $X_1, ..., X_n$ samples. My best guess at the correct definition of $V^2$ is $$V^2 := frac1nsum_i=1^n (X_i-mu)^2$$
– Michael
Jul 15 at 10:45
Agree with @Michael. // If $hattau$ is estimator of $tau,$ the $MSE(hat tau) = E[(tau - hat tau)^2|.$ Also, $MSE(hat tau) = Var(hat tau),$ provided that $E(hattau) = tau$ (that is provided that $hat tau$ is unbiased). // Finally, for $X_1 sim mathsfNorm(mu, sigma)$ one has $left(fracX_1 - musigmaright)^2 sim mathsfChisq(1).$ You can look up info on mean and variance of chi-squared dist'n on Wikipleda or your text.
– BruceET
Jul 15 at 19:03
Except for the bad start defining $V^2$ incorrectly, you have pretty much the right idea. @Michael showed you how to start. Hope my Answ gives you clues how to fix the derivation.
– BruceET
Jul 15 at 23:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $ X_1,...,X_n$ ~ $N(mu, sigma^2)$ where $mu$ is known and $sigma^2$ is unknown, calculate the MSE of $V^2$
$V^2 = frac1n sum_X_i^n Var(X_i) =sigma^2$
Therefore:
$MSE(V^2) = Var(V^2) = frac1n^2nVar[(X_1-mu)^2]=frac1nVar[sigma^2(fracX_1-musigma)^2]=frac1nsigma^4Var[(fracX_1-musigma)^2]=frac2sigma^4n$
However, I do not understand some of the steps:
- Where does the $X_1$ suddenly come from (instead of$ X_i$)?
- And then in the next step, I am aware it has somehing to do with the fact that $fracX-musigma$ ~ $chi^2_1$ But i cannot connect the dots .
Could someone break these down for me ? I do not have an mathematical background, therefore stating the obvious is very welcome.
statistics convergence parameter-estimation mean-square-error
asked Jul 15 at 9:43
Danka
178
If $ X_1,...,X_n$ ~ $N(mu, sigma^2)$ where $mu$ is known and $sigma^2$ is unknown, calculate the MSE of $V^2$
$V^2 = frac1n sum_X_i^n Var(X_i) =sigma^2$
Therefore:
$MSE(V^2) = Var(V^2) = frac1n^2nVar[(X_1-mu)^2]=frac1nVar[sigma^2(fracX_1-musigma)^2]=frac1nsigma^4Var[(fracX_1-musigma)^2]=frac2sigma^4n$
However, I do not understand some of the steps:
- Where does the $X_1$ suddenly come from (instead of$ X_i$)?
- And then in the next step, I am aware it has somehing to do with the fact that $fracX-musigma$ ~ $chi^2_1$ But i cannot connect the dots .
Could someone break these down for me ? I do not have an mathematical background, therefore stating the obvious is very welcome.
statistics convergence parameter-estimation mean-square-error
asked Jul 15 at 9:43
Danka
178
asked Jul 15 at 9:43
Danka
178
asked Jul 15 at 9:43
Danka
178
asked Jul 15 at 9:43
Danka
178
178
closed as unclear what you're asking by StubbornAtom, Mostafa Ayaz, José Carlos Santos, user223391, Parcly Taxel Jul 16 at 4:26
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by StubbornAtom, Mostafa Ayaz, José Carlos Santos, user223391, Parcly Taxel Jul 16 at 4:26
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
If $X_i sim N(mu, sigma^2)$ then what is the difference between $Var(X_i)$ and $sigma^2$? Are the $X_i$ independent? Also, what is the meaning of $sum_X_i^n$? Finally, it looks like you are defining $V^2$ as the constant $sigma^2$ so there is no estimation going on and the variance of $V^2$ is 0. I suspect you are incorrectly interpreting a problem you are given, I would expect $V^2$ to be some estimate formed from the $X_1, ..., X_n$ samples. My best guess at the correct definition of $V^2$ is $$V^2 := frac1nsum_i=1^n (X_i-mu)^2$$
– Michael
Jul 15 at 10:45
Agree with @Michael. // If $hattau$ is estimator of $tau,$ the $MSE(hat tau) = E[(tau - hat tau)^2|.$ Also, $MSE(hat tau) = Var(hat tau),$ provided that $E(hattau) = tau$ (that is provided that $hat tau$ is unbiased). // Finally, for $X_1 sim mathsfNorm(mu, sigma)$ one has $left(fracX_1 - musigmaright)^2 sim mathsfChisq(1).$ You can look up info on mean and variance of chi-squared dist'n on Wikipleda or your text.
– BruceET
Jul 15 at 19:03
Except for the bad start defining $V^2$ incorrectly, you have pretty much the right idea. @Michael showed you how to start. Hope my Answ gives you clues how to fix the derivation.
– BruceET
Jul 15 at 23:10
add a comment |Â
2
If $X_i sim N(mu, sigma^2)$ then what is the difference between $Var(X_i)$ and $sigma^2$? Are the $X_i$ independent? Also, what is the meaning of $sum_X_i^n$? Finally, it looks like you are defining $V^2$ as the constant $sigma^2$ so there is no estimation going on and the variance of $V^2$ is 0. I suspect you are incorrectly interpreting a problem you are given, I would expect $V^2$ to be some estimate formed from the $X_1, ..., X_n$ samples. My best guess at the correct definition of $V^2$ is $$V^2 := frac1nsum_i=1^n (X_i-mu)^2$$
– Michael
Jul 15 at 10:45
Agree with @Michael. // If $hattau$ is estimator of $tau,$ the $MSE(hat tau) = E[(tau - hat tau)^2|.$ Also, $MSE(hat tau) = Var(hat tau),$ provided that $E(hattau) = tau$ (that is provided that $hat tau$ is unbiased). // Finally, for $X_1 sim mathsfNorm(mu, sigma)$ one has $left(fracX_1 - musigmaright)^2 sim mathsfChisq(1).$ You can look up info on mean and variance of chi-squared dist'n on Wikipleda or your text.
– BruceET
Jul 15 at 19:03
Except for the bad start defining $V^2$ incorrectly, you have pretty much the right idea. @Michael showed you how to start. Hope my Answ gives you clues how to fix the derivation.
– BruceET
Jul 15 at 23:10
2
2
If $X_i sim N(mu, sigma^2)$ then what is the difference between $Var(X_i)$ and $sigma^2$? Are the $X_i$ independent? Also, what is the meaning of $sum_X_i^n$? Finally, it looks like you are defining $V^2$ as the constant $sigma^2$ so there is no estimation going on and the variance of $V^2$ is 0. I suspect you are incorrectly interpreting a problem you are given, I would expect $V^2$ to be some estimate formed from the $X_1, ..., X_n$ samples. My best guess at the correct definition of $V^2$ is $$V^2 := frac1nsum_i=1^n (X_i-mu)^2$$
– Michael
Jul 15 at 10:45
If $X_i sim N(mu, sigma^2)$ then what is the difference between $Var(X_i)$ and $sigma^2$? Are the $X_i$ independent? Also, what is the meaning of $sum_X_i^n$? Finally, it looks like you are defining $V^2$ as the constant $sigma^2$ so there is no estimation going on and the variance of $V^2$ is 0. I suspect you are incorrectly interpreting a problem you are given, I would expect $V^2$ to be some estimate formed from the $X_1, ..., X_n$ samples. My best guess at the correct definition of $V^2$ is $$V^2 := frac1nsum_i=1^n (X_i-mu)^2$$
– Michael
Jul 15 at 10:45
Agree with @Michael. // If $hattau$ is estimator of $tau,$ the $MSE(hat tau) = E[(tau - hat tau)^2|.$ Also, $MSE(hat tau) = Var(hat tau),$ provided that $E(hattau) = tau$ (that is provided that $hat tau$ is unbiased). // Finally, for $X_1 sim mathsfNorm(mu, sigma)$ one has $left(fracX_1 - musigmaright)^2 sim mathsfChisq(1).$ You can look up info on mean and variance of chi-squared dist'n on Wikipleda or your text.
– BruceET
Jul 15 at 19:03
Agree with @Michael. // If $hattau$ is estimator of $tau,$ the $MSE(hat tau) = E[(tau - hat tau)^2|.$ Also, $MSE(hat tau) = Var(hat tau),$ provided that $E(hattau) = tau$ (that is provided that $hat tau$ is unbiased). // Finally, for $X_1 sim mathsfNorm(mu, sigma)$ one has $left(fracX_1 - musigmaright)^2 sim mathsfChisq(1).$ You can look up info on mean and variance of chi-squared dist'n on Wikipleda or your text.
– BruceET
Jul 15 at 19:03
Except for the bad start defining $V^2$ incorrectly, you have pretty much the right idea. @Michael showed you how to start. Hope my Answ gives you clues how to fix the derivation.
– BruceET
Jul 15 at 23:10
Except for the bad start defining $V^2$ incorrectly, you have pretty much the right idea. @Michael showed you how to start. Hope my Answ gives you clues how to fix the derivation.
– BruceET
Jul 15 at 23:10
add a comment |Â
1 Answer
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up vote
1
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accepted
Let $X_1, X_2, dots X_n$ be a random sample from $mathsfNorm(mu, sigma),$ where $mu$ is known and $sigma^2$ is to be estimated by
$V = frac 1 nsum_i=1^n (X_i - mu)^2.$ (Note the use of the known population mean $mu,$ not the sample mean $bar X.)$ You want to evaluate
$MSE(V).$ @Michael and I have given you some hints. (Notice that my $V$ is
your $V^2$ to simplify notation a bit.)
With that orientation, I hope the following example with specific numbers for the quantities involved will help you do the required general derivation.
Suppose $n = 5,, mu = 0$ and $sigma = 4.$ Then
$Q = fracnVsigma^2 = frac5V16 sim mathsfChisq(n=5),$ which has mean $n=5$ and variance $2n=10.$ So $E(V) = fracsigma^2nn = 16,$ (showing that $V$ is unbiased for $sigma^2)$ and
$Var(V) = MSE(V) = fracsigma^4n^22n = 102.4.$
The following demonstration, using R statistical software, with a million such samples of size $n=5$
illustrates these numerical results to several significant digits. In
the program MAT is a $10^6 times 5$ matrix, in which each row is a sample
of size $5.$
set.seed(715) # retain for exactly same simulation, delete for fresh run
m = 10^6; n = 5; mu = 0; sg = 4
x = rnorm(m*n, mu, sg); MAT = matrix(x, nrow = m)
v = rowMeans((MAT - mu)^2) # using 'known' population mean, not sample mean
mean(v); mean((v-sg^2)^2)
[1] 15.99998 # aprx E(V) = 16
[1] 102.5 # aprs MSE(V) = 102.4
The plot below shows the simulated distribution of
$Q = fracnVsigma^2 = frac5V16 = 0.3125V$ along with the density curve of $mathsfChisq(5).$
hist(5*v/sg^2, prob=T, br=40, xlab="q", col="skyblue2", main="")
curve(dchisq(x, 5), add=T, lwd=2, n=1001)
answered Jul 15 at 22:52
BruceET
33.3k61440
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $X_1, X_2, dots X_n$ be a random sample from $mathsfNorm(mu, sigma),$ where $mu$ is known and $sigma^2$ is to be estimated by
$V = frac 1 nsum_i=1^n (X_i - mu)^2.$ (Note the use of the known population mean $mu,$ not the sample mean $bar X.)$ You want to evaluate
$MSE(V).$ @Michael and I have given you some hints. (Notice that my $V$ is
your $V^2$ to simplify notation a bit.)
With that orientation, I hope the following example with specific numbers for the quantities involved will help you do the required general derivation.
Suppose $n = 5,, mu = 0$ and $sigma = 4.$ Then
$Q = fracnVsigma^2 = frac5V16 sim mathsfChisq(n=5),$ which has mean $n=5$ and variance $2n=10.$ So $E(V) = fracsigma^2nn = 16,$ (showing that $V$ is unbiased for $sigma^2)$ and
$Var(V) = MSE(V) = fracsigma^4n^22n = 102.4.$
The following demonstration, using R statistical software, with a million such samples of size $n=5$
illustrates these numerical results to several significant digits. In
the program MAT is a $10^6 times 5$ matrix, in which each row is a sample
of size $5.$
set.seed(715) # retain for exactly same simulation, delete for fresh run
m = 10^6; n = 5; mu = 0; sg = 4
x = rnorm(m*n, mu, sg); MAT = matrix(x, nrow = m)
v = rowMeans((MAT - mu)^2) # using 'known' population mean, not sample mean
mean(v); mean((v-sg^2)^2)
[1] 15.99998 # aprx E(V) = 16
[1] 102.5 # aprs MSE(V) = 102.4
The plot below shows the simulated distribution of
$Q = fracnVsigma^2 = frac5V16 = 0.3125V$ along with the density curve of $mathsfChisq(5).$
hist(5*v/sg^2, prob=T, br=40, xlab="q", col="skyblue2", main="")
curve(dchisq(x, 5), add=T, lwd=2, n=1001)
answered Jul 15 at 22:52
BruceET
33.3k61440
add a comment |Â
up vote
1
down vote
accepted
Let $X_1, X_2, dots X_n$ be a random sample from $mathsfNorm(mu, sigma),$ where $mu$ is known and $sigma^2$ is to be estimated by
$V = frac 1 nsum_i=1^n (X_i - mu)^2.$ (Note the use of the known population mean $mu,$ not the sample mean $bar X.)$ You want to evaluate
$MSE(V).$ @Michael and I have given you some hints. (Notice that my $V$ is
your $V^2$ to simplify notation a bit.)
With that orientation, I hope the following example with specific numbers for the quantities involved will help you do the required general derivation.
Suppose $n = 5,, mu = 0$ and $sigma = 4.$ Then
$Q = fracnVsigma^2 = frac5V16 sim mathsfChisq(n=5),$ which has mean $n=5$ and variance $2n=10.$ So $E(V) = fracsigma^2nn = 16,$ (showing that $V$ is unbiased for $sigma^2)$ and
$Var(V) = MSE(V) = fracsigma^4n^22n = 102.4.$
The following demonstration, using R statistical software, with a million such samples of size $n=5$
illustrates these numerical results to several significant digits. In
the program MAT is a $10^6 times 5$ matrix, in which each row is a sample
of size $5.$
set.seed(715) # retain for exactly same simulation, delete for fresh run
m = 10^6; n = 5; mu = 0; sg = 4
x = rnorm(m*n, mu, sg); MAT = matrix(x, nrow = m)
v = rowMeans((MAT - mu)^2) # using 'known' population mean, not sample mean
mean(v); mean((v-sg^2)^2)
[1] 15.99998 # aprx E(V) = 16
[1] 102.5 # aprs MSE(V) = 102.4
The plot below shows the simulated distribution of
$Q = fracnVsigma^2 = frac5V16 = 0.3125V$ along with the density curve of $mathsfChisq(5).$
hist(5*v/sg^2, prob=T, br=40, xlab="q", col="skyblue2", main="")
curve(dchisq(x, 5), add=T, lwd=2, n=1001)
answered Jul 15 at 22:52
BruceET
33.3k61440
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $X_1, X_2, dots X_n$ be a random sample from $mathsfNorm(mu, sigma),$ where $mu$ is known and $sigma^2$ is to be estimated by
$V = frac 1 nsum_i=1^n (X_i - mu)^2.$ (Note the use of the known population mean $mu,$ not the sample mean $bar X.)$ You want to evaluate
$MSE(V).$ @Michael and I have given you some hints. (Notice that my $V$ is
your $V^2$ to simplify notation a bit.)
With that orientation, I hope the following example with specific numbers for the quantities involved will help you do the required general derivation.
Suppose $n = 5,, mu = 0$ and $sigma = 4.$ Then
$Q = fracnVsigma^2 = frac5V16 sim mathsfChisq(n=5),$ which has mean $n=5$ and variance $2n=10.$ So $E(V) = fracsigma^2nn = 16,$ (showing that $V$ is unbiased for $sigma^2)$ and
$Var(V) = MSE(V) = fracsigma^4n^22n = 102.4.$
The following demonstration, using R statistical software, with a million such samples of size $n=5$
illustrates these numerical results to several significant digits. In
the program MAT is a $10^6 times 5$ matrix, in which each row is a sample
of size $5.$
set.seed(715) # retain for exactly same simulation, delete for fresh run
m = 10^6; n = 5; mu = 0; sg = 4
x = rnorm(m*n, mu, sg); MAT = matrix(x, nrow = m)
v = rowMeans((MAT - mu)^2) # using 'known' population mean, not sample mean
mean(v); mean((v-sg^2)^2)
[1] 15.99998 # aprx E(V) = 16
[1] 102.5 # aprs MSE(V) = 102.4
The plot below shows the simulated distribution of
$Q = fracnVsigma^2 = frac5V16 = 0.3125V$ along with the density curve of $mathsfChisq(5).$
hist(5*v/sg^2, prob=T, br=40, xlab="q", col="skyblue2", main="")
curve(dchisq(x, 5), add=T, lwd=2, n=1001)
answered Jul 15 at 22:52
BruceET
33.3k61440
Let $X_1, X_2, dots X_n$ be a random sample from $mathsfNorm(mu, sigma),$ where $mu$ is known and $sigma^2$ is to be estimated by
$V = frac 1 nsum_i=1^n (X_i - mu)^2.$ (Note the use of the known population mean $mu,$ not the sample mean $bar X.)$ You want to evaluate
$MSE(V).$ @Michael and I have given you some hints. (Notice that my $V$ is
your $V^2$ to simplify notation a bit.)
With that orientation, I hope the following example with specific numbers for the quantities involved will help you do the required general derivation.
Suppose $n = 5,, mu = 0$ and $sigma = 4.$ Then
$Q = fracnVsigma^2 = frac5V16 sim mathsfChisq(n=5),$ which has mean $n=5$ and variance $2n=10.$ So $E(V) = fracsigma^2nn = 16,$ (showing that $V$ is unbiased for $sigma^2)$ and
$Var(V) = MSE(V) = fracsigma^4n^22n = 102.4.$
The following demonstration, using R statistical software, with a million such samples of size $n=5$
illustrates these numerical results to several significant digits. In
the program MAT is a $10^6 times 5$ matrix, in which each row is a sample
of size $5.$
set.seed(715) # retain for exactly same simulation, delete for fresh run
m = 10^6; n = 5; mu = 0; sg = 4
x = rnorm(m*n, mu, sg); MAT = matrix(x, nrow = m)
v = rowMeans((MAT - mu)^2) # using 'known' population mean, not sample mean
mean(v); mean((v-sg^2)^2)
[1] 15.99998 # aprx E(V) = 16
[1] 102.5 # aprs MSE(V) = 102.4
The plot below shows the simulated distribution of
$Q = fracnVsigma^2 = frac5V16 = 0.3125V$ along with the density curve of $mathsfChisq(5).$
hist(5*v/sg^2, prob=T, br=40, xlab="q", col="skyblue2", main="")
curve(dchisq(x, 5), add=T, lwd=2, n=1001)
answered Jul 15 at 22:52
BruceET
33.3k61440
edited Jul 15 at 23:36
answered Jul 15 at 22:52
BruceET
33.3k61440
answered Jul 15 at 22:52
BruceET
33.3k61440
answered Jul 15 at 22:52
BruceET
33.3k61440
33.3k61440
add a comment |Â
add a comment |Â
2
If $X_i sim N(mu, sigma^2)$ then what is the difference between $Var(X_i)$ and $sigma^2$? Are the $X_i$ independent? Also, what is the meaning of $sum_X_i^n$? Finally, it looks like you are defining $V^2$ as the constant $sigma^2$ so there is no estimation going on and the variance of $V^2$ is 0. I suspect you are incorrectly interpreting a problem you are given, I would expect $V^2$ to be some estimate formed from the $X_1, ..., X_n$ samples. My best guess at the correct definition of $V^2$ is $$V^2 := frac1nsum_i=1^n (X_i-mu)^2$$
– Michael
Jul 15 at 10:45
Agree with @Michael. // If $hattau$ is estimator of $tau,$ the $MSE(hat tau) = E[(tau - hat tau)^2|.$ Also, $MSE(hat tau) = Var(hat tau),$ provided that $E(hattau) = tau$ (that is provided that $hat tau$ is unbiased). // Finally, for $X_1 sim mathsfNorm(mu, sigma)$ one has $left(fracX_1 - musigmaright)^2 sim mathsfChisq(1).$ You can look up info on mean and variance of chi-squared dist'n on Wikipleda or your text.
– BruceET
Jul 15 at 19:03
Except for the bad start defining $V^2$ incorrectly, you have pretty much the right idea. @Michael showed you how to start. Hope my Answ gives you clues how to fix the derivation.
– BruceET
Jul 15 at 23:10