Solving CLT with Exponential Distributions
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I am working on a CLT problem and am a bit stuck.
The problem:
N1 = 10,000 with claims X1 ~ Exp() with mean = 100
N2 = 3000 with claims X2 ~ Exp() with mean = 200
N3 = 1000 with claims X3 ~ Exp() with mean 1000
All independent, find the Value at Risk for 99%
What I have done so far:
To find the $lambda$ for each of the exponential distributions from the mean:
E[X] = mean and E[X] = $1overlambda$ for Exp()
So...
X1 ~ Exp($lambda$) = $1over lambda$ = 100 so $lambda$ = .01
X2 ~ Exp($lambda$) = $1over lambda$ = 200 so $lambda$ = .005
X3 ~ Exp($lambda$) = $1over lambda$ = 1000 so $lambda$ = .001
With this...
E[X1] = 100 / E[X2] = 200 / E[X3] = 1000
Var[X1] = $1over (.01)^2$ = 10,000
Var[X2] = $1over (.005)^2$ = 40,000
Var[X3] = $1over (.001)^2$ = 1,000,000
Then...
E[S] = 100(N1) + 200(N2) + 1000(N3) = (100)(10,000) + (200)(3000) + (1000)(1000) = 2,600,000
Var[S] = 10,000(N1) + 40,000(N2) + 1,000,000(N3) = (10,000)(10,000) + (40,000)(3000) + (1,000,000)(1000) = 10,220,000,000
So using the CLT then...
$ S - 2,600,000 over sqrt 10,220,000,000$ $le$ 2.326
Is this the correct approach? These numbers seem a little bit off to me.
probability probability-theory probability-distributions central-limit-theorem
asked Dec 3 '18 at 18:47
EthanEthan
10513
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add a comment |
$begingroup$
I am working on a CLT problem and am a bit stuck.
The problem:
N1 = 10,000 with claims X1 ~ Exp() with mean = 100
N2 = 3000 with claims X2 ~ Exp() with mean = 200
N3 = 1000 with claims X3 ~ Exp() with mean 1000
All independent, find the Value at Risk for 99%
What I have done so far:
To find the $lambda$ for each of the exponential distributions from the mean:
E[X] = mean and E[X] = $1overlambda$ for Exp()
So...
X1 ~ Exp($lambda$) = $1over lambda$ = 100 so $lambda$ = .01
X2 ~ Exp($lambda$) = $1over lambda$ = 200 so $lambda$ = .005
X3 ~ Exp($lambda$) = $1over lambda$ = 1000 so $lambda$ = .001
With this...
E[X1] = 100 / E[X2] = 200 / E[X3] = 1000
Var[X1] = $1over (.01)^2$ = 10,000
Var[X2] = $1over (.005)^2$ = 40,000
Var[X3] = $1over (.001)^2$ = 1,000,000
Then...
E[S] = 100(N1) + 200(N2) + 1000(N3) = (100)(10,000) + (200)(3000) + (1000)(1000) = 2,600,000
Var[S] = 10,000(N1) + 40,000(N2) + 1,000,000(N3) = (10,000)(10,000) + (40,000)(3000) + (1,000,000)(1000) = 10,220,000,000
So using the CLT then...
$ S - 2,600,000 over sqrt 10,220,000,000$ $le$ 2.326
Is this the correct approach? These numbers seem a little bit off to me.
probability probability-theory probability-distributions central-limit-theorem
asked Dec 3 '18 at 18:47
EthanEthan
10513
$endgroup$
add a comment |
$begingroup$
I am working on a CLT problem and am a bit stuck.
The problem:
N1 = 10,000 with claims X1 ~ Exp() with mean = 100
N2 = 3000 with claims X2 ~ Exp() with mean = 200
N3 = 1000 with claims X3 ~ Exp() with mean 1000
All independent, find the Value at Risk for 99%
What I have done so far:
To find the $lambda$ for each of the exponential distributions from the mean:
E[X] = mean and E[X] = $1overlambda$ for Exp()
So...
X1 ~ Exp($lambda$) = $1over lambda$ = 100 so $lambda$ = .01
X2 ~ Exp($lambda$) = $1over lambda$ = 200 so $lambda$ = .005
X3 ~ Exp($lambda$) = $1over lambda$ = 1000 so $lambda$ = .001
With this...
E[X1] = 100 / E[X2] = 200 / E[X3] = 1000
Var[X1] = $1over (.01)^2$ = 10,000
Var[X2] = $1over (.005)^2$ = 40,000
Var[X3] = $1over (.001)^2$ = 1,000,000
Then...
E[S] = 100(N1) + 200(N2) + 1000(N3) = (100)(10,000) + (200)(3000) + (1000)(1000) = 2,600,000
Var[S] = 10,000(N1) + 40,000(N2) + 1,000,000(N3) = (10,000)(10,000) + (40,000)(3000) + (1,000,000)(1000) = 10,220,000,000
So using the CLT then...
$ S - 2,600,000 over sqrt 10,220,000,000$ $le$ 2.326
Is this the correct approach? These numbers seem a little bit off to me.
probability probability-theory probability-distributions central-limit-theorem
asked Dec 3 '18 at 18:47
EthanEthan
10513
$endgroup$
I am working on a CLT problem and am a bit stuck.
The problem:
N1 = 10,000 with claims X1 ~ Exp() with mean = 100
N2 = 3000 with claims X2 ~ Exp() with mean = 200
N3 = 1000 with claims X3 ~ Exp() with mean 1000
All independent, find the Value at Risk for 99%
What I have done so far:
To find the $lambda$ for each of the exponential distributions from the mean:
E[X] = mean and E[X] = $1overlambda$ for Exp()
So...
X1 ~ Exp($lambda$) = $1over lambda$ = 100 so $lambda$ = .01
X2 ~ Exp($lambda$) = $1over lambda$ = 200 so $lambda$ = .005
X3 ~ Exp($lambda$) = $1over lambda$ = 1000 so $lambda$ = .001
With this...
E[X1] = 100 / E[X2] = 200 / E[X3] = 1000
Var[X1] = $1over (.01)^2$ = 10,000
Var[X2] = $1over (.005)^2$ = 40,000
Var[X3] = $1over (.001)^2$ = 1,000,000
Then...
E[S] = 100(N1) + 200(N2) + 1000(N3) = (100)(10,000) + (200)(3000) + (1000)(1000) = 2,600,000
Var[S] = 10,000(N1) + 40,000(N2) + 1,000,000(N3) = (10,000)(10,000) + (40,000)(3000) + (1,000,000)(1000) = 10,220,000,000
So using the CLT then...
$ S - 2,600,000 over sqrt 10,220,000,000$ $le$ 2.326
Is this the correct approach? These numbers seem a little bit off to me.
probability probability-theory probability-distributions central-limit-theorem
probability probability-theory probability-distributions central-limit-theorem
asked Dec 3 '18 at 18:47
EthanEthan
10513
asked Dec 3 '18 at 18:47
EthanEthan
10513
edited Dec 4 '18 at 1:57
Ethan
asked Dec 3 '18 at 18:47
EthanEthan
10513
asked Dec 3 '18 at 18:47
EthanEthan
10513
asked Dec 3 '18 at 18:47
EthanEthan
10513
10513
add a comment |
add a comment |
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