Gumble distribution probability
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If I want to find y such that P(Y< y)=0.999 of Gumbel Distribution with
a=30.134711, b= 7.621868.
where $$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
Is it correct that I use quantile of Gumbel distribution with p=0.999 to find such y which is =82.7809
statistics statistical-inference
asked Nov 18 at 22:07
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up vote
0
down vote
favorite
If I want to find y such that P(Y< y)=0.999 of Gumbel Distribution with
a=30.134711, b= 7.621868.
where $$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
Is it correct that I use quantile of Gumbel distribution with p=0.999 to find such y which is =82.7809
statistics statistical-inference
asked Nov 18 at 22:07
Extra mint
124
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I want to find y such that P(Y< y)=0.999 of Gumbel Distribution with
a=30.134711, b= 7.621868.
where $$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
Is it correct that I use quantile of Gumbel distribution with p=0.999 to find such y which is =82.7809
statistics statistical-inference
asked Nov 18 at 22:07
Extra mint
124
If I want to find y such that P(Y< y)=0.999 of Gumbel Distribution with
a=30.134711, b= 7.621868.
where $$f(y,;a,b)=frac{1}{b}expleft[-frac{(y-a)}{b}-e^{-(y-a)/b}right]quad,,yinmathbb Rquad,,ainmathbb R,b>0$$
Is it correct that I use quantile of Gumbel distribution with p=0.999 to find such y which is =82.7809
statistics statistical-inference
statistics statistical-inference
asked Nov 18 at 22:07
Extra mint
124
asked Nov 18 at 22:07
Extra mint
124
asked Nov 18 at 22:07
Extra mint
124
asked Nov 18 at 22:07
Extra mint
124
asked Nov 18 at 22:07
Extra mint
124
124
add a comment |
add a comment |
1 Answer
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Yes, that looks correct
From that density, you have $$mathbb P(Y le y)=F(y)= exp(-exp(-(y-a)/b))$$
and inverting this gives $$F^{-1}(p)= a-blog_e(-log_e(p))$$
then substituting $a=30.134711, b= 7.621868, p=0.999$ gives about $y=82.7809$
answered Nov 18 at 23:21
Henry
96.7k474154
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yes, that looks correct
From that density, you have $$mathbb P(Y le y)=F(y)= exp(-exp(-(y-a)/b))$$
and inverting this gives $$F^{-1}(p)= a-blog_e(-log_e(p))$$
then substituting $a=30.134711, b= 7.621868, p=0.999$ gives about $y=82.7809$
answered Nov 18 at 23:21
Henry
96.7k474154
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
add a comment |
up vote
0
down vote
accepted
Yes, that looks correct
From that density, you have $$mathbb P(Y le y)=F(y)= exp(-exp(-(y-a)/b))$$
and inverting this gives $$F^{-1}(p)= a-blog_e(-log_e(p))$$
then substituting $a=30.134711, b= 7.621868, p=0.999$ gives about $y=82.7809$
answered Nov 18 at 23:21
Henry
96.7k474154
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes, that looks correct
From that density, you have $$mathbb P(Y le y)=F(y)= exp(-exp(-(y-a)/b))$$
and inverting this gives $$F^{-1}(p)= a-blog_e(-log_e(p))$$
then substituting $a=30.134711, b= 7.621868, p=0.999$ gives about $y=82.7809$
answered Nov 18 at 23:21
Henry
96.7k474154
Yes, that looks correct
From that density, you have $$mathbb P(Y le y)=F(y)= exp(-exp(-(y-a)/b))$$
and inverting this gives $$F^{-1}(p)= a-blog_e(-log_e(p))$$
then substituting $a=30.134711, b= 7.621868, p=0.999$ gives about $y=82.7809$
answered Nov 18 at 23:21
Henry
96.7k474154
answered Nov 18 at 23:21
Henry
96.7k474154
answered Nov 18 at 23:21
Henry
96.7k474154
answered Nov 18 at 23:21
Henry
96.7k474154
96.7k474154
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
add a comment |
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
Thank you sir.Another question follow from that is this equivalent to that “estimate the daily rainfall total that will be exceeded at least once in 2019 with probability 0.001”
– Extra mint
Nov 18 at 23:38
add a comment |
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