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Derivation from “The concrete distribution: A continuous relaxation of discrete random variablesâ€
Clash Royale CLAN TAG#URR8PPP
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I am looking at "The concrete distribution: A continuous relaxation of discrete random variables".
https://arxiv.org/pdf/1611.00712.pdf
I do not understand the step "integrating our $r$" on page $13$ in Appendix A. Please let me know if you know why this step is true.
Thank you.
probability probability-distributions machine-learning data-analysis
asked Jul 26 at 1:20
Roger
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up vote
-1
down vote
favorite
I am looking at "The concrete distribution: A continuous relaxation of discrete random variables".
https://arxiv.org/pdf/1611.00712.pdf
I do not understand the step "integrating our $r$" on page $13$ in Appendix A. Please let me know if you know why this step is true.
Thank you.
probability probability-distributions machine-learning data-analysis
asked Jul 26 at 1:20
Roger
536
They are using this definition of $Gamma(n)$, in which the integral had the change of variable $x=e^-lambda r+gamma$ and $r$ the new integration variable.
– user578878
Jul 26 at 2:03
@nextpuzzle why definition, could you please explain more in detail? thx
– Roger
Jul 26 at 2:06
$Gamma(n)$ is just a function with that integral in the link as definition. Make the change of variable I said and you will see how the integral becomes the integral of the factors that depends on $r$ in your article, on the appropriate interval given by the change of variable.
– user578878
Jul 26 at 2:08
@nextpuzzle I tried. Honestly, I did not even get why there is even an integral, as the previous step has no integral at all...
– Roger
Jul 26 at 2:15
They are not saying that the two expressions in that step are equal. They are saying that they are integrating with respect to $r$.
– user578878
Jul 26 at 2:20
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I am looking at "The concrete distribution: A continuous relaxation of discrete random variables".
https://arxiv.org/pdf/1611.00712.pdf
I do not understand the step "integrating our $r$" on page $13$ in Appendix A. Please let me know if you know why this step is true.
Thank you.
probability probability-distributions machine-learning data-analysis
asked Jul 26 at 1:20
Roger
536
I am looking at "The concrete distribution: A continuous relaxation of discrete random variables".
https://arxiv.org/pdf/1611.00712.pdf
I do not understand the step "integrating our $r$" on page $13$ in Appendix A. Please let me know if you know why this step is true.
Thank you.
probability probability-distributions machine-learning data-analysis
asked Jul 26 at 1:20
Roger
536
edited Jul 26 at 10:05
asked Jul 26 at 1:20
Roger
536
asked Jul 26 at 1:20
Roger
536
asked Jul 26 at 1:20
Roger
536
536
They are using this definition of $Gamma(n)$, in which the integral had the change of variable $x=e^-lambda r+gamma$ and $r$ the new integration variable.
– user578878
Jul 26 at 2:03
@nextpuzzle why definition, could you please explain more in detail? thx
– Roger
Jul 26 at 2:06
$Gamma(n)$ is just a function with that integral in the link as definition. Make the change of variable I said and you will see how the integral becomes the integral of the factors that depends on $r$ in your article, on the appropriate interval given by the change of variable.
– user578878
Jul 26 at 2:08
@nextpuzzle I tried. Honestly, I did not even get why there is even an integral, as the previous step has no integral at all...
– Roger
Jul 26 at 2:15
They are not saying that the two expressions in that step are equal. They are saying that they are integrating with respect to $r$.
– user578878
Jul 26 at 2:20
 |Â
show 1 more comment
They are using this definition of $Gamma(n)$, in which the integral had the change of variable $x=e^-lambda r+gamma$ and $r$ the new integration variable.
– user578878
Jul 26 at 2:03
@nextpuzzle why definition, could you please explain more in detail? thx
– Roger
Jul 26 at 2:06
$Gamma(n)$ is just a function with that integral in the link as definition. Make the change of variable I said and you will see how the integral becomes the integral of the factors that depends on $r$ in your article, on the appropriate interval given by the change of variable.
– user578878
Jul 26 at 2:08
@nextpuzzle I tried. Honestly, I did not even get why there is even an integral, as the previous step has no integral at all...
– Roger
Jul 26 at 2:15
They are not saying that the two expressions in that step are equal. They are saying that they are integrating with respect to $r$.
– user578878
Jul 26 at 2:20
They are using this definition of $Gamma(n)$, in which the integral had the change of variable $x=e^-lambda r+gamma$ and $r$ the new integration variable.
– user578878
Jul 26 at 2:03
They are using this definition of $Gamma(n)$, in which the integral had the change of variable $x=e^-lambda r+gamma$ and $r$ the new integration variable.
– user578878
Jul 26 at 2:03
@nextpuzzle why definition, could you please explain more in detail? thx
– Roger
Jul 26 at 2:06
@nextpuzzle why definition, could you please explain more in detail? thx
– Roger
Jul 26 at 2:06
$Gamma(n)$ is just a function with that integral in the link as definition. Make the change of variable I said and you will see how the integral becomes the integral of the factors that depends on $r$ in your article, on the appropriate interval given by the change of variable.
– user578878
Jul 26 at 2:08
$Gamma(n)$ is just a function with that integral in the link as definition. Make the change of variable I said and you will see how the integral becomes the integral of the factors that depends on $r$ in your article, on the appropriate interval given by the change of variable.
– user578878
Jul 26 at 2:08
@nextpuzzle I tried. Honestly, I did not even get why there is even an integral, as the previous step has no integral at all...
– Roger
Jul 26 at 2:15
@nextpuzzle I tried. Honestly, I did not even get why there is even an integral, as the previous step has no integral at all...
– Roger
Jul 26 at 2:15
They are not saying that the two expressions in that step are equal. They are saying that they are integrating with respect to $r$.
– user578878
Jul 26 at 2:20
They are not saying that the two expressions in that step are equal. They are saying that they are integrating with respect to $r$.
– user578878
Jul 26 at 2:20
 |Â
show 1 more comment
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They are using this definition of $Gamma(n)$, in which the integral had the change of variable $x=e^-lambda r+gamma$ and $r$ the new integration variable.
– user578878
Jul 26 at 2:03
@nextpuzzle why definition, could you please explain more in detail? thx
– Roger
Jul 26 at 2:06
$Gamma(n)$ is just a function with that integral in the link as definition. Make the change of variable I said and you will see how the integral becomes the integral of the factors that depends on $r$ in your article, on the appropriate interval given by the change of variable.
– user578878
Jul 26 at 2:08
@nextpuzzle I tried. Honestly, I did not even get why there is even an integral, as the previous step has no integral at all...
– Roger
Jul 26 at 2:15
They are not saying that the two expressions in that step are equal. They are saying that they are integrating with respect to $r$.
– user578878
Jul 26 at 2:20