Mixing
Pivots for exponential distribution
up vote
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In my mathematical statistics course I got the following problem:
Let $X_{1}, ... , X_{n}$ be an i.i.d. sample from the Exp($lambda$) distribution. Construct two different pivots
and two confidence intervals for $lambda$ (of confidence level $1-alpha$) based on these pivots. Use that if $X sim Exp(lambda) Rightarrow lambda X sim Exp(1)$ in combination with the following two facts (do not prove them):
(1) $X_{(1)}$ $sim$ Exp($nlambda$),
(2) if $Y_{1},...,Y_{n}$ are i.i.d. with Exp($1$) distribution, then $2sumlimits_{i=1}^{n}Y_{i} sim chi^{2}_{2n}$.
So far this is what I got:
We will first use $2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n}$. We now use $y=lambda x$, so we get
begin{equation*}
begin{split}
2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n} &Rightarrow 2 lambda sumlimits_{i=1}^{n} x_{i} = 2 lambda bar{x} n sim chi^{2}_{2n}\
end{split}
end{equation*}
Thus
begin{equation*}
begin{split}
chi^{2}_{2n,alpha / 2} &leq 2lambda bar{x} n leq chi^{2}_{2n, 1-alpha /2}\
frac{chi^{2}_{2n,alpha / 2}}{2 bar{x} n} &leq lambda leq frac{chi^{2}_{2n, 1-alpha /2}}{2 bar{x} n}
end{split}
end{equation*}
So this is my first pivot. I have no idea whether this is okay, and I also have no clue how to proceed for the second pivot. Any suggestions are welcome!
statistics parameter-estimation
asked 2 days ago
Mathbeginner
1328
add a comment |
up vote
1
down vote
favorite
In my mathematical statistics course I got the following problem:
Let $X_{1}, ... , X_{n}$ be an i.i.d. sample from the Exp($lambda$) distribution. Construct two different pivots
and two confidence intervals for $lambda$ (of confidence level $1-alpha$) based on these pivots. Use that if $X sim Exp(lambda) Rightarrow lambda X sim Exp(1)$ in combination with the following two facts (do not prove them):
(1) $X_{(1)}$ $sim$ Exp($nlambda$),
(2) if $Y_{1},...,Y_{n}$ are i.i.d. with Exp($1$) distribution, then $2sumlimits_{i=1}^{n}Y_{i} sim chi^{2}_{2n}$.
So far this is what I got:
We will first use $2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n}$. We now use $y=lambda x$, so we get
begin{equation*}
begin{split}
2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n} &Rightarrow 2 lambda sumlimits_{i=1}^{n} x_{i} = 2 lambda bar{x} n sim chi^{2}_{2n}\
end{split}
end{equation*}
Thus
begin{equation*}
begin{split}
chi^{2}_{2n,alpha / 2} &leq 2lambda bar{x} n leq chi^{2}_{2n, 1-alpha /2}\
frac{chi^{2}_{2n,alpha / 2}}{2 bar{x} n} &leq lambda leq frac{chi^{2}_{2n, 1-alpha /2}}{2 bar{x} n}
end{split}
end{equation*}
So this is my first pivot. I have no idea whether this is okay, and I also have no clue how to proceed for the second pivot. Any suggestions are welcome!
statistics parameter-estimation
asked 2 days ago
Mathbeginner
1328
Using your notation, the second pivot could be $nlambda X_{(1)}sim text{Exp}(1)$ or equivalently, $2nlambda X_{(1)}simchi^2_2$.
– StubbornAtom
yesterday
Thanks, the first one makes sense to me. The second one, not so much, but that must be because this is all still a bit of a vague area for me.
– Mathbeginner
yesterday
1
A Chi-square variable with 2 degrees of freedom is just an Exponential variable with mean 2.
– StubbornAtom
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my mathematical statistics course I got the following problem:
Let $X_{1}, ... , X_{n}$ be an i.i.d. sample from the Exp($lambda$) distribution. Construct two different pivots
and two confidence intervals for $lambda$ (of confidence level $1-alpha$) based on these pivots. Use that if $X sim Exp(lambda) Rightarrow lambda X sim Exp(1)$ in combination with the following two facts (do not prove them):
(1) $X_{(1)}$ $sim$ Exp($nlambda$),
(2) if $Y_{1},...,Y_{n}$ are i.i.d. with Exp($1$) distribution, then $2sumlimits_{i=1}^{n}Y_{i} sim chi^{2}_{2n}$.
So far this is what I got:
We will first use $2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n}$. We now use $y=lambda x$, so we get
begin{equation*}
begin{split}
2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n} &Rightarrow 2 lambda sumlimits_{i=1}^{n} x_{i} = 2 lambda bar{x} n sim chi^{2}_{2n}\
end{split}
end{equation*}
Thus
begin{equation*}
begin{split}
chi^{2}_{2n,alpha / 2} &leq 2lambda bar{x} n leq chi^{2}_{2n, 1-alpha /2}\
frac{chi^{2}_{2n,alpha / 2}}{2 bar{x} n} &leq lambda leq frac{chi^{2}_{2n, 1-alpha /2}}{2 bar{x} n}
end{split}
end{equation*}
So this is my first pivot. I have no idea whether this is okay, and I also have no clue how to proceed for the second pivot. Any suggestions are welcome!
statistics parameter-estimation
asked 2 days ago
Mathbeginner
1328
In my mathematical statistics course I got the following problem:
Let $X_{1}, ... , X_{n}$ be an i.i.d. sample from the Exp($lambda$) distribution. Construct two different pivots
and two confidence intervals for $lambda$ (of confidence level $1-alpha$) based on these pivots. Use that if $X sim Exp(lambda) Rightarrow lambda X sim Exp(1)$ in combination with the following two facts (do not prove them):
(1) $X_{(1)}$ $sim$ Exp($nlambda$),
(2) if $Y_{1},...,Y_{n}$ are i.i.d. with Exp($1$) distribution, then $2sumlimits_{i=1}^{n}Y_{i} sim chi^{2}_{2n}$.
So far this is what I got:
We will first use $2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n}$. We now use $y=lambda x$, so we get
begin{equation*}
begin{split}
2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n} &Rightarrow 2 lambda sumlimits_{i=1}^{n} x_{i} = 2 lambda bar{x} n sim chi^{2}_{2n}\
end{split}
end{equation*}
Thus
begin{equation*}
begin{split}
chi^{2}_{2n,alpha / 2} &leq 2lambda bar{x} n leq chi^{2}_{2n, 1-alpha /2}\
frac{chi^{2}_{2n,alpha / 2}}{2 bar{x} n} &leq lambda leq frac{chi^{2}_{2n, 1-alpha /2}}{2 bar{x} n}
end{split}
end{equation*}
So this is my first pivot. I have no idea whether this is okay, and I also have no clue how to proceed for the second pivot. Any suggestions are welcome!
statistics parameter-estimation
statistics parameter-estimation
asked 2 days ago
Mathbeginner
1328
asked 2 days ago
Mathbeginner
1328
asked 2 days ago
Mathbeginner
1328
asked 2 days ago
Mathbeginner
1328
asked 2 days ago
Mathbeginner
1328
1328
Using your notation, the second pivot could be $nlambda X_{(1)}sim text{Exp}(1)$ or equivalently, $2nlambda X_{(1)}simchi^2_2$.
– StubbornAtom
yesterday
Thanks, the first one makes sense to me. The second one, not so much, but that must be because this is all still a bit of a vague area for me.
– Mathbeginner
yesterday
1
A Chi-square variable with 2 degrees of freedom is just an Exponential variable with mean 2.
– StubbornAtom
yesterday
add a comment |
Using your notation, the second pivot could be $nlambda X_{(1)}sim text{Exp}(1)$ or equivalently, $2nlambda X_{(1)}simchi^2_2$.
– StubbornAtom
yesterday
Thanks, the first one makes sense to me. The second one, not so much, but that must be because this is all still a bit of a vague area for me.
– Mathbeginner
yesterday
1
A Chi-square variable with 2 degrees of freedom is just an Exponential variable with mean 2.
– StubbornAtom
yesterday
Using your notation, the second pivot could be $nlambda X_{(1)}sim text{Exp}(1)$ or equivalently, $2nlambda X_{(1)}simchi^2_2$.
– StubbornAtom
yesterday
Using your notation, the second pivot could be $nlambda X_{(1)}sim text{Exp}(1)$ or equivalently, $2nlambda X_{(1)}simchi^2_2$.
– StubbornAtom
yesterday
Thanks, the first one makes sense to me. The second one, not so much, but that must be because this is all still a bit of a vague area for me.
– Mathbeginner
yesterday
Thanks, the first one makes sense to me. The second one, not so much, but that must be because this is all still a bit of a vague area for me.
– Mathbeginner
yesterday
1
1
A Chi-square variable with 2 degrees of freedom is just an Exponential variable with mean 2.
– StubbornAtom
yesterday
A Chi-square variable with 2 degrees of freedom is just an Exponential variable with mean 2.
– StubbornAtom
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your derivation of the confidence interval for rate $lambda$ using the sample mean
and the chi-squared distribution is correct. You can verify the result at Wikipedia under 'Confidence intervals'. This is a commonly used CI for $lambda$ using the sufficient statistic.
The derivation of the CI for $lambda$ using the minimum $X_{(1)} = V$ is similar.
The quantity $nVlambda sim mathsf{Exp}(1).$ For example,
$$P(L_e < nVlambda < U_e) = Pleft(frac{L_e}{nV} < lambda < frac{U_e}{nV}right) = 0.95,$$
where $L_e$ and $U_e$ cut probability 0.25 from the lower and upper tails of
$mathsf{Exp}(1),$ respectively.
Below is a simulation in R with a million iterations, for $n = 20, lambda = 0.2.$
m = 10^6; n = 20; lam = .2; x = rexp(m*n, lam)
MAT = matrix(x, nrow=m) # m x n matrix, each row a sample of size n
s = rowSums(MAT) # one million sample sums
mean(s); sd(s)
[1] 100.0436 # aprx E(S) = 100
[1] 22.36627
L.c = qchisq(.025,2*n); U.c = qchisq(.975,2*n); L.c; U.c
[1] 24.43304
[1] 59.34171
mean(L.c/(2*s) < lam & U.c/(2*s) > lam)
[1] 0.950114 # CI covers rate with probability 95%
v = apply(MAT, 1, min) # one million sample minimums
mean(v); sd(v)
[1] 0.2501739 # aprx E(V) = 1/4
[1] 0.2503302
L.e = qexp(.025); U.e = qexp(.975); L.e; U.e
[1] 0.02531781
[1] 3.688879
mean(L.e/(n*v) < lam & U.e/(n*v) > lam)
[1] 0.950174 # CI covers rate with probability 95%
The histogram in the left panel below shows the simulated chi-squared distribution of $2nbar Xlambda = 2lambdasum_iX_i$ and the right panel
shows the simulated exponential distribution of $nVlambda = nlambda X_{(1)}.$
Exact density functions are shown as black curves.
R code to make the figure follows:
par(mfrow=c(1,2))
hist(2*s*lam, prob=T, col="skyblue2", main="Pivot Using Sum")
curve(dchisq(x,2*n), add=T, lwd=2); abline(v=c(L.c, U.c), col="red")
hist(n*lam*v, prob=T, ylim=0:1, col="skyblue2", main="Pivot Using Minimum")
curve(dexp(x), add=T, lwd=2); abline(v=c(L.e, U.e), col="red")
par(mfrow=c(1,1))
answered yesterday
BruceET
34.7k71440
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your derivation of the confidence interval for rate $lambda$ using the sample mean
and the chi-squared distribution is correct. You can verify the result at Wikipedia under 'Confidence intervals'. This is a commonly used CI for $lambda$ using the sufficient statistic.
The derivation of the CI for $lambda$ using the minimum $X_{(1)} = V$ is similar.
The quantity $nVlambda sim mathsf{Exp}(1).$ For example,
$$P(L_e < nVlambda < U_e) = Pleft(frac{L_e}{nV} < lambda < frac{U_e}{nV}right) = 0.95,$$
where $L_e$ and $U_e$ cut probability 0.25 from the lower and upper tails of
$mathsf{Exp}(1),$ respectively.
Below is a simulation in R with a million iterations, for $n = 20, lambda = 0.2.$
m = 10^6; n = 20; lam = .2; x = rexp(m*n, lam)
MAT = matrix(x, nrow=m) # m x n matrix, each row a sample of size n
s = rowSums(MAT) # one million sample sums
mean(s); sd(s)
[1] 100.0436 # aprx E(S) = 100
[1] 22.36627
L.c = qchisq(.025,2*n); U.c = qchisq(.975,2*n); L.c; U.c
[1] 24.43304
[1] 59.34171
mean(L.c/(2*s) < lam & U.c/(2*s) > lam)
[1] 0.950114 # CI covers rate with probability 95%
v = apply(MAT, 1, min) # one million sample minimums
mean(v); sd(v)
[1] 0.2501739 # aprx E(V) = 1/4
[1] 0.2503302
L.e = qexp(.025); U.e = qexp(.975); L.e; U.e
[1] 0.02531781
[1] 3.688879
mean(L.e/(n*v) < lam & U.e/(n*v) > lam)
[1] 0.950174 # CI covers rate with probability 95%
The histogram in the left panel below shows the simulated chi-squared distribution of $2nbar Xlambda = 2lambdasum_iX_i$ and the right panel
shows the simulated exponential distribution of $nVlambda = nlambda X_{(1)}.$
Exact density functions are shown as black curves.
R code to make the figure follows:
par(mfrow=c(1,2))
hist(2*s*lam, prob=T, col="skyblue2", main="Pivot Using Sum")
curve(dchisq(x,2*n), add=T, lwd=2); abline(v=c(L.c, U.c), col="red")
hist(n*lam*v, prob=T, ylim=0:1, col="skyblue2", main="Pivot Using Minimum")
curve(dexp(x), add=T, lwd=2); abline(v=c(L.e, U.e), col="red")
par(mfrow=c(1,1))
answered yesterday
BruceET
34.7k71440
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
add a comment |
up vote
1
down vote
accepted
Your derivation of the confidence interval for rate $lambda$ using the sample mean
and the chi-squared distribution is correct. You can verify the result at Wikipedia under 'Confidence intervals'. This is a commonly used CI for $lambda$ using the sufficient statistic.
The derivation of the CI for $lambda$ using the minimum $X_{(1)} = V$ is similar.
The quantity $nVlambda sim mathsf{Exp}(1).$ For example,
$$P(L_e < nVlambda < U_e) = Pleft(frac{L_e}{nV} < lambda < frac{U_e}{nV}right) = 0.95,$$
where $L_e$ and $U_e$ cut probability 0.25 from the lower and upper tails of
$mathsf{Exp}(1),$ respectively.
Below is a simulation in R with a million iterations, for $n = 20, lambda = 0.2.$
m = 10^6; n = 20; lam = .2; x = rexp(m*n, lam)
MAT = matrix(x, nrow=m) # m x n matrix, each row a sample of size n
s = rowSums(MAT) # one million sample sums
mean(s); sd(s)
[1] 100.0436 # aprx E(S) = 100
[1] 22.36627
L.c = qchisq(.025,2*n); U.c = qchisq(.975,2*n); L.c; U.c
[1] 24.43304
[1] 59.34171
mean(L.c/(2*s) < lam & U.c/(2*s) > lam)
[1] 0.950114 # CI covers rate with probability 95%
v = apply(MAT, 1, min) # one million sample minimums
mean(v); sd(v)
[1] 0.2501739 # aprx E(V) = 1/4
[1] 0.2503302
L.e = qexp(.025); U.e = qexp(.975); L.e; U.e
[1] 0.02531781
[1] 3.688879
mean(L.e/(n*v) < lam & U.e/(n*v) > lam)
[1] 0.950174 # CI covers rate with probability 95%
The histogram in the left panel below shows the simulated chi-squared distribution of $2nbar Xlambda = 2lambdasum_iX_i$ and the right panel
shows the simulated exponential distribution of $nVlambda = nlambda X_{(1)}.$
Exact density functions are shown as black curves.
R code to make the figure follows:
par(mfrow=c(1,2))
hist(2*s*lam, prob=T, col="skyblue2", main="Pivot Using Sum")
curve(dchisq(x,2*n), add=T, lwd=2); abline(v=c(L.c, U.c), col="red")
hist(n*lam*v, prob=T, ylim=0:1, col="skyblue2", main="Pivot Using Minimum")
curve(dexp(x), add=T, lwd=2); abline(v=c(L.e, U.e), col="red")
par(mfrow=c(1,1))
answered yesterday
BruceET
34.7k71440
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your derivation of the confidence interval for rate $lambda$ using the sample mean
and the chi-squared distribution is correct. You can verify the result at Wikipedia under 'Confidence intervals'. This is a commonly used CI for $lambda$ using the sufficient statistic.
The derivation of the CI for $lambda$ using the minimum $X_{(1)} = V$ is similar.
The quantity $nVlambda sim mathsf{Exp}(1).$ For example,
$$P(L_e < nVlambda < U_e) = Pleft(frac{L_e}{nV} < lambda < frac{U_e}{nV}right) = 0.95,$$
where $L_e$ and $U_e$ cut probability 0.25 from the lower and upper tails of
$mathsf{Exp}(1),$ respectively.
Below is a simulation in R with a million iterations, for $n = 20, lambda = 0.2.$
m = 10^6; n = 20; lam = .2; x = rexp(m*n, lam)
MAT = matrix(x, nrow=m) # m x n matrix, each row a sample of size n
s = rowSums(MAT) # one million sample sums
mean(s); sd(s)
[1] 100.0436 # aprx E(S) = 100
[1] 22.36627
L.c = qchisq(.025,2*n); U.c = qchisq(.975,2*n); L.c; U.c
[1] 24.43304
[1] 59.34171
mean(L.c/(2*s) < lam & U.c/(2*s) > lam)
[1] 0.950114 # CI covers rate with probability 95%
v = apply(MAT, 1, min) # one million sample minimums
mean(v); sd(v)
[1] 0.2501739 # aprx E(V) = 1/4
[1] 0.2503302
L.e = qexp(.025); U.e = qexp(.975); L.e; U.e
[1] 0.02531781
[1] 3.688879
mean(L.e/(n*v) < lam & U.e/(n*v) > lam)
[1] 0.950174 # CI covers rate with probability 95%
The histogram in the left panel below shows the simulated chi-squared distribution of $2nbar Xlambda = 2lambdasum_iX_i$ and the right panel
shows the simulated exponential distribution of $nVlambda = nlambda X_{(1)}.$
Exact density functions are shown as black curves.
R code to make the figure follows:
par(mfrow=c(1,2))
hist(2*s*lam, prob=T, col="skyblue2", main="Pivot Using Sum")
curve(dchisq(x,2*n), add=T, lwd=2); abline(v=c(L.c, U.c), col="red")
hist(n*lam*v, prob=T, ylim=0:1, col="skyblue2", main="Pivot Using Minimum")
curve(dexp(x), add=T, lwd=2); abline(v=c(L.e, U.e), col="red")
par(mfrow=c(1,1))
answered yesterday
BruceET
34.7k71440
Your derivation of the confidence interval for rate $lambda$ using the sample mean
and the chi-squared distribution is correct. You can verify the result at Wikipedia under 'Confidence intervals'. This is a commonly used CI for $lambda$ using the sufficient statistic.
The derivation of the CI for $lambda$ using the minimum $X_{(1)} = V$ is similar.
The quantity $nVlambda sim mathsf{Exp}(1).$ For example,
$$P(L_e < nVlambda < U_e) = Pleft(frac{L_e}{nV} < lambda < frac{U_e}{nV}right) = 0.95,$$
where $L_e$ and $U_e$ cut probability 0.25 from the lower and upper tails of
$mathsf{Exp}(1),$ respectively.
Below is a simulation in R with a million iterations, for $n = 20, lambda = 0.2.$
m = 10^6; n = 20; lam = .2; x = rexp(m*n, lam)
MAT = matrix(x, nrow=m) # m x n matrix, each row a sample of size n
s = rowSums(MAT) # one million sample sums
mean(s); sd(s)
[1] 100.0436 # aprx E(S) = 100
[1] 22.36627
L.c = qchisq(.025,2*n); U.c = qchisq(.975,2*n); L.c; U.c
[1] 24.43304
[1] 59.34171
mean(L.c/(2*s) < lam & U.c/(2*s) > lam)
[1] 0.950114 # CI covers rate with probability 95%
v = apply(MAT, 1, min) # one million sample minimums
mean(v); sd(v)
[1] 0.2501739 # aprx E(V) = 1/4
[1] 0.2503302
L.e = qexp(.025); U.e = qexp(.975); L.e; U.e
[1] 0.02531781
[1] 3.688879
mean(L.e/(n*v) < lam & U.e/(n*v) > lam)
[1] 0.950174 # CI covers rate with probability 95%
The histogram in the left panel below shows the simulated chi-squared distribution of $2nbar Xlambda = 2lambdasum_iX_i$ and the right panel
shows the simulated exponential distribution of $nVlambda = nlambda X_{(1)}.$
Exact density functions are shown as black curves.
R code to make the figure follows:
par(mfrow=c(1,2))
hist(2*s*lam, prob=T, col="skyblue2", main="Pivot Using Sum")
curve(dchisq(x,2*n), add=T, lwd=2); abline(v=c(L.c, U.c), col="red")
hist(n*lam*v, prob=T, ylim=0:1, col="skyblue2", main="Pivot Using Minimum")
curve(dexp(x), add=T, lwd=2); abline(v=c(L.e, U.e), col="red")
par(mfrow=c(1,1))
answered yesterday
BruceET
34.7k71440
edited yesterday
answered yesterday
BruceET
34.7k71440
answered yesterday
BruceET
34.7k71440
answered yesterday
BruceET
34.7k71440
34.7k71440
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
add a comment |
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Very nice, thanks a lot. Really appreciate the amount of work, the graphs make it a lot more intuitive!
– Mathbeginner
9 hours ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
Can you figure out how to find the average length of each kind of CI? Of course, shorter is better.
– BruceET
1 hour ago
add a comment |
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Using your notation, the second pivot could be $nlambda X_{(1)}sim text{Exp}(1)$ or equivalently, $2nlambda X_{(1)}simchi^2_2$.
– StubbornAtom
yesterday
Thanks, the first one makes sense to me. The second one, not so much, but that must be because this is all still a bit of a vague area for me.
– Mathbeginner
yesterday
1
A Chi-square variable with 2 degrees of freedom is just an Exponential variable with mean 2.
– StubbornAtom
yesterday