What does the Squared 2-Parameter Exponential Cumulative Density Function Measure?

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The 2-parameter exponential cumulative density function is defined as $1-e^lambda (x-gamma)quad$ (see e.g. this page )

Question:

what does $quad1-left(1-e^lambda (x-gamma)right)^2quad$ measure?

Any information about that distribution would be appreciated.

probability-distributions

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edited Jul 27 at 19:49

asked Jul 27 at 19:17

Manfred Weis

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up vote
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favorite

The 2-parameter exponential cumulative density function is defined as $1-e^lambda (x-gamma)quad$ (see e.g. this page )

Question:

what does $quad1-left(1-e^lambda (x-gamma)right)^2quad$ measure?

Any information about that distribution would be appreciated.

probability-distributions

share|cite|improve this question

edited Jul 27 at 19:49

asked Jul 27 at 19:17

Manfred Weis

1417

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

The 2-parameter exponential cumulative density function is defined as $1-e^lambda (x-gamma)quad$ (see e.g. this page )

Question:

what does $quad1-left(1-e^lambda (x-gamma)right)^2quad$ measure?

Any information about that distribution would be appreciated.

probability-distributions

share|cite|improve this question

edited Jul 27 at 19:49

asked Jul 27 at 19:17

Manfred Weis

1417

The 2-parameter exponential cumulative density function is defined as $1-e^lambda (x-gamma)quad$ (see e.g. this page )

Question:

what does $quad1-left(1-e^lambda (x-gamma)right)^2quad$ measure?

Any information about that distribution would be appreciated.

probability-distributions

share|cite|improve this question

edited Jul 27 at 19:49

asked Jul 27 at 19:17

Manfred Weis

1417

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 19:49

edited Jul 27 at 19:49

edited Jul 27 at 19:49

asked Jul 27 at 19:17

Manfred Weis

1417

asked Jul 27 at 19:17

Manfred Weis

1417

asked Jul 27 at 19:17

Manfred Weis

1417

1417

add a comment | 

add a comment | 

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$1-e^lambda (x-gamma)$ is $mathbb P(Xle x)$ for a two parameter exponentially distributed random variable with rate $lambda$ and minimum possible value $gamma$ where $xge gamma$.

Suppose you have two independent random variables $X_1$ and $X_2$ with this distribution. Then $left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(X_1le x, X_2le x) = mathbb P(max(X_1, X_2)le x)$. So $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x)$

Alternatively, if $X_1$ and $X_2$ have random variables with independent one parameter exponential distributions and rates of $lambda$, then $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x mid min(X_1, X_2)gt gamma)$ so long as $0 le gamma le x$

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answered Jul 27 at 21:27

Henry

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1 Answer
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1 Answer
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up vote
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$1-e^lambda (x-gamma)$ is $mathbb P(Xle x)$ for a two parameter exponentially distributed random variable with rate $lambda$ and minimum possible value $gamma$ where $xge gamma$.

Suppose you have two independent random variables $X_1$ and $X_2$ with this distribution. Then $left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(X_1le x, X_2le x) = mathbb P(max(X_1, X_2)le x)$. So $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x)$

Alternatively, if $X_1$ and $X_2$ have random variables with independent one parameter exponential distributions and rates of $lambda$, then $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x mid min(X_1, X_2)gt gamma)$ so long as $0 le gamma le x$

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answered Jul 27 at 21:27

Henry

92.9k469147

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up vote
1
down vote

$1-e^lambda (x-gamma)$ is $mathbb P(Xle x)$ for a two parameter exponentially distributed random variable with rate $lambda$ and minimum possible value $gamma$ where $xge gamma$.

Suppose you have two independent random variables $X_1$ and $X_2$ with this distribution. Then $left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(X_1le x, X_2le x) = mathbb P(max(X_1, X_2)le x)$. So $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x)$

Alternatively, if $X_1$ and $X_2$ have random variables with independent one parameter exponential distributions and rates of $lambda$, then $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x mid min(X_1, X_2)gt gamma)$ so long as $0 le gamma le x$

share|cite|improve this answer

answered Jul 27 at 21:27

Henry

92.9k469147

add a comment | 

up vote
1
down vote

up vote
1
down vote

$1-e^lambda (x-gamma)$ is $mathbb P(Xle x)$ for a two parameter exponentially distributed random variable with rate $lambda$ and minimum possible value $gamma$ where $xge gamma$.

Suppose you have two independent random variables $X_1$ and $X_2$ with this distribution. Then $left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(X_1le x, X_2le x) = mathbb P(max(X_1, X_2)le x)$. So $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x)$

Alternatively, if $X_1$ and $X_2$ have random variables with independent one parameter exponential distributions and rates of $lambda$, then $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x mid min(X_1, X_2)gt gamma)$ so long as $0 le gamma le x$

share|cite|improve this answer

answered Jul 27 at 21:27

Henry

92.9k469147

$1-e^lambda (x-gamma)$ is $mathbb P(Xle x)$ for a two parameter exponentially distributed random variable with rate $lambda$ and minimum possible value $gamma$ where $xge gamma$.

Suppose you have two independent random variables $X_1$ and $X_2$ with this distribution. Then $left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(X_1le x, X_2le x) = mathbb P(max(X_1, X_2)le x)$. So $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x)$

Alternatively, if $X_1$ and $X_2$ have random variables with independent one parameter exponential distributions and rates of $lambda$, then $1-left(1-e^lambda (x-gamma)right)^2$ is $mathbb P(max(X_1, X_2)gt x mid min(X_1, X_2)gt gamma)$ so long as $0 le gamma le x$

share|cite|improve this answer

answered Jul 27 at 21:27

Henry

92.9k469147

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 21:27

Henry

92.9k469147

answered Jul 27 at 21:27

Henry

92.9k469147

answered Jul 27 at 21:27

Henry

92.9k469147

92.9k469147

add a comment | 

add a comment | 

 

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