Create random variable Expo from Unif(0,1)
$begingroup$
Working through some problems from Introduction to Probability (Blitzstein)
Let U~Unif(0,1). Using U, construct X~Expo($lambda$).
My work:
(edited with updates on CDF and inverse function)
PDF of X is $lambda e^{-lambda x}$ by definition of exponential
distributionCDF is therefore $1-e^{-lambda x}$
Use Universality of the Uniform: find inverse function for F
$F^{-1}(x) = frac{ln(1-x)}{-lambda}$
( $y=1-e^{-lambda x} -> 1-y = e^{-lambda x}) -> x = frac{ln(1-y)}{-lambda} $)
- $F^{-1}(U) = frac{ln(1-U)}{-lambda}$
I am pretty confused by the process of getting a random variable from a Uniform distribution. Any help?
random-variables uniform-distribution
asked Dec 2 '18 at 16:31
user603569user603569
728
$endgroup$
add a comment |
$begingroup$
Working through some problems from Introduction to Probability (Blitzstein)
Let U~Unif(0,1). Using U, construct X~Expo($lambda$).
My work:
(edited with updates on CDF and inverse function)
PDF of X is $lambda e^{-lambda x}$ by definition of exponential
distributionCDF is therefore $1-e^{-lambda x}$
Use Universality of the Uniform: find inverse function for F
$F^{-1}(x) = frac{ln(1-x)}{-lambda}$
( $y=1-e^{-lambda x} -> 1-y = e^{-lambda x}) -> x = frac{ln(1-y)}{-lambda} $)
- $F^{-1}(U) = frac{ln(1-U)}{-lambda}$
I am pretty confused by the process of getting a random variable from a Uniform distribution. Any help?
random-variables uniform-distribution
asked Dec 2 '18 at 16:31
user603569user603569
728
$endgroup$
$begingroup$
Your CDF and your $F^{-1}$ are incorrect. Please revise your work with much more care.
$endgroup$
– Did
Dec 2 '18 at 16:41
$begingroup$
A very big red flag should be the fact that you're getting negative values for the CDF!
$endgroup$
– Boshu
Dec 2 '18 at 17:18
add a comment |
$begingroup$
Working through some problems from Introduction to Probability (Blitzstein)
Let U~Unif(0,1). Using U, construct X~Expo($lambda$).
My work:
(edited with updates on CDF and inverse function)
PDF of X is $lambda e^{-lambda x}$ by definition of exponential
distributionCDF is therefore $1-e^{-lambda x}$
Use Universality of the Uniform: find inverse function for F
$F^{-1}(x) = frac{ln(1-x)}{-lambda}$
( $y=1-e^{-lambda x} -> 1-y = e^{-lambda x}) -> x = frac{ln(1-y)}{-lambda} $)
- $F^{-1}(U) = frac{ln(1-U)}{-lambda}$
I am pretty confused by the process of getting a random variable from a Uniform distribution. Any help?
random-variables uniform-distribution
asked Dec 2 '18 at 16:31
user603569user603569
728
$endgroup$
Working through some problems from Introduction to Probability (Blitzstein)
Let U~Unif(0,1). Using U, construct X~Expo($lambda$).
My work:
(edited with updates on CDF and inverse function)
PDF of X is $lambda e^{-lambda x}$ by definition of exponential
distributionCDF is therefore $1-e^{-lambda x}$
Use Universality of the Uniform: find inverse function for F
$F^{-1}(x) = frac{ln(1-x)}{-lambda}$
( $y=1-e^{-lambda x} -> 1-y = e^{-lambda x}) -> x = frac{ln(1-y)}{-lambda} $)
- $F^{-1}(U) = frac{ln(1-U)}{-lambda}$
I am pretty confused by the process of getting a random variable from a Uniform distribution. Any help?
random-variables uniform-distribution
random-variables uniform-distribution
asked Dec 2 '18 at 16:31
user603569user603569
728
asked Dec 2 '18 at 16:31
user603569user603569
728
edited Dec 3 '18 at 11:43
user603569
asked Dec 2 '18 at 16:31
user603569user603569
728
asked Dec 2 '18 at 16:31
user603569user603569
728
asked Dec 2 '18 at 16:31
user603569user603569
728
728
$begingroup$
Your CDF and your $F^{-1}$ are incorrect. Please revise your work with much more care.
$endgroup$
– Did
Dec 2 '18 at 16:41
$begingroup$
A very big red flag should be the fact that you're getting negative values for the CDF!
$endgroup$
– Boshu
Dec 2 '18 at 17:18
add a comment |
$begingroup$
Your CDF and your $F^{-1}$ are incorrect. Please revise your work with much more care.
$endgroup$
– Did
Dec 2 '18 at 16:41
$begingroup$
A very big red flag should be the fact that you're getting negative values for the CDF!
$endgroup$
– Boshu
Dec 2 '18 at 17:18
$begingroup$
Your CDF and your $F^{-1}$ are incorrect. Please revise your work with much more care.
$endgroup$
– Did
Dec 2 '18 at 16:41
$begingroup$
Your CDF and your $F^{-1}$ are incorrect. Please revise your work with much more care.
$endgroup$
– Did
Dec 2 '18 at 16:41
$begingroup$
A very big red flag should be the fact that you're getting negative values for the CDF!
$endgroup$
– Boshu
Dec 2 '18 at 17:18
$begingroup$
A very big red flag should be the fact that you're getting negative values for the CDF!
$endgroup$
– Boshu
Dec 2 '18 at 17:18
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
So there's a mistake in your CDF as has been pointed out (hint: you need to add something) and I'll leave it to you to correct subsequent errors. If you want to know why this method works, then consider the fact that the CDF is a mapping of the support on the interval $[0,1]$. Therefore, if you have a random variable $Usimtext{Unif}[0,1]$, then the random variable $Z=F^{-1}_X(U)$ has the distribution $F_X$. So what you're doing is essentially this: you have the graph of the CDF of your distribution of interest. You pick a number $xin[0,1]$ and you see where the CDF attains this value. This is your simulated random variable, and your $x$ is generated by the uniform distribution.
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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votes
active
oldest
votes
$begingroup$
So there's a mistake in your CDF as has been pointed out (hint: you need to add something) and I'll leave it to you to correct subsequent errors. If you want to know why this method works, then consider the fact that the CDF is a mapping of the support on the interval $[0,1]$. Therefore, if you have a random variable $Usimtext{Unif}[0,1]$, then the random variable $Z=F^{-1}_X(U)$ has the distribution $F_X$. So what you're doing is essentially this: you have the graph of the CDF of your distribution of interest. You pick a number $xin[0,1]$ and you see where the CDF attains this value. This is your simulated random variable, and your $x$ is generated by the uniform distribution.
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
$endgroup$
add a comment |
$begingroup$
So there's a mistake in your CDF as has been pointed out (hint: you need to add something) and I'll leave it to you to correct subsequent errors. If you want to know why this method works, then consider the fact that the CDF is a mapping of the support on the interval $[0,1]$. Therefore, if you have a random variable $Usimtext{Unif}[0,1]$, then the random variable $Z=F^{-1}_X(U)$ has the distribution $F_X$. So what you're doing is essentially this: you have the graph of the CDF of your distribution of interest. You pick a number $xin[0,1]$ and you see where the CDF attains this value. This is your simulated random variable, and your $x$ is generated by the uniform distribution.
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
$endgroup$
add a comment |
$begingroup$
So there's a mistake in your CDF as has been pointed out (hint: you need to add something) and I'll leave it to you to correct subsequent errors. If you want to know why this method works, then consider the fact that the CDF is a mapping of the support on the interval $[0,1]$. Therefore, if you have a random variable $Usimtext{Unif}[0,1]$, then the random variable $Z=F^{-1}_X(U)$ has the distribution $F_X$. So what you're doing is essentially this: you have the graph of the CDF of your distribution of interest. You pick a number $xin[0,1]$ and you see where the CDF attains this value. This is your simulated random variable, and your $x$ is generated by the uniform distribution.
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
$endgroup$
So there's a mistake in your CDF as has been pointed out (hint: you need to add something) and I'll leave it to you to correct subsequent errors. If you want to know why this method works, then consider the fact that the CDF is a mapping of the support on the interval $[0,1]$. Therefore, if you have a random variable $Usimtext{Unif}[0,1]$, then the random variable $Z=F^{-1}_X(U)$ has the distribution $F_X$. So what you're doing is essentially this: you have the graph of the CDF of your distribution of interest. You pick a number $xin[0,1]$ and you see where the CDF attains this value. This is your simulated random variable, and your $x$ is generated by the uniform distribution.
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
answered Dec 2 '18 at 17:09
BoshuBoshu
703315
703315
add a comment |
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$begingroup$
Your CDF and your $F^{-1}$ are incorrect. Please revise your work with much more care.
$endgroup$
– Did
Dec 2 '18 at 16:41
$begingroup$
A very big red flag should be the fact that you're getting negative values for the CDF!
$endgroup$
– Boshu
Dec 2 '18 at 17:18