Unnormalized Log Probability - RNN












2












$begingroup$


I am going through the deep learning book by Goodfellow. In the RNN section I am stuck with the following:



RNN is defined like following:



enter image description here



And the equations are :



enter image description here



enter image description here



Now the $O^{(t)}$ above is considered as unnormalized log probability. But if this is true, then the value of $O^{(t)}$ must be negative because,



Probability is always defined as a number between 0 and 1, i.e, $Pin[0,1]$, where brackets denote closed interval. And $log(P) le 0$ on this interval. But in the equations above, nowhere this condition that $O^{(t)} le 0$ is explicitly enforced.



What am I missing!






deep-learning lstm recurrent-neural-net







share|improve this question











$endgroup$

















    2












    $begingroup$


    I am going through the deep learning book by Goodfellow. In the RNN section I am stuck with the following:



    RNN is defined like following:



    enter image description here



    And the equations are :



    enter image description here



    enter image description here



    Now the $O^{(t)}$ above is considered as unnormalized log probability. But if this is true, then the value of $O^{(t)}$ must be negative because,



    Probability is always defined as a number between 0 and 1, i.e, $Pin[0,1]$, where brackets denote closed interval. And $log(P) le 0$ on this interval. But in the equations above, nowhere this condition that $O^{(t)} le 0$ is explicitly enforced.



    What am I missing!






    deep-learning lstm recurrent-neural-net







    share|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am going through the deep learning book by Goodfellow. In the RNN section I am stuck with the following:



      RNN is defined like following:



      enter image description here



      And the equations are :



      enter image description here



      enter image description here



      Now the $O^{(t)}$ above is considered as unnormalized log probability. But if this is true, then the value of $O^{(t)}$ must be negative because,



      Probability is always defined as a number between 0 and 1, i.e, $Pin[0,1]$, where brackets denote closed interval. And $log(P) le 0$ on this interval. But in the equations above, nowhere this condition that $O^{(t)} le 0$ is explicitly enforced.



      What am I missing!






      deep-learning lstm recurrent-neural-net







      share|improve this question











      $endgroup$




      I am going through the deep learning book by Goodfellow. In the RNN section I am stuck with the following:



      RNN is defined like following:



      enter image description here



      And the equations are :



      enter image description here



      enter image description here



      Now the $O^{(t)}$ above is considered as unnormalized log probability. But if this is true, then the value of $O^{(t)}$ must be negative because,



      Probability is always defined as a number between 0 and 1, i.e, $Pin[0,1]$, where brackets denote closed interval. And $log(P) le 0$ on this interval. But in the equations above, nowhere this condition that $O^{(t)} le 0$ is explicitly enforced.



      What am I missing!






      deep-learning lstm recurrent-neural-net




      deep-learning lstm recurrent-neural-net






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited yesterday









      Siong Thye Goh

      1,322418




      1,322418










      asked yesterday









      user3001408user3001408

      360146




      360146






















          2 Answers
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          $begingroup$

          You are right in a sense that it is better to be called log of unnormalized probability. This way, the quantity could be positive or negative. For example, $text{log}(0.5) < 0$ and $text{log}(12) > 0$ are both valid log of unnormalized probabilities. Here, in more detail:




          1. Probability: $P(i) = e^{o_i}/sum_{k=1}^{K}e^{o_k}$ (using softmax as mentioned in Figure 10.3 caption, and assuming $mathbf{o}=(o_1,..,o_K)$ is the output of layer before softmax),


          2. Unnormalized probability: $tilde{P}(i) = e^{o_i}$, which can be larger than 1,


          3. Log of unnormalized probability: $text{log}tilde{P}(i) = o_i$, which can be positive or negative.







          share|improve this answer











          $endgroup$





















            2












            $begingroup$

            You are right, nothing stop $o_k^{(t)}$ from being nonnegative, the keyword here is "unnormalized".



            If we let $o_k^{(t)}=ln q_k^{(t)}$



            $$hat{y}^{(t)}_k= frac{exp(o^{(t)}_k)}{sum_{k=1}^K exp(o^{(t)}_k) }= frac{q_k^{(t)}}{sum_{k=1}^K q_k^{(t)}}$$



            Here $q_k^{(t)}$ can be any positive number, they will be normalized to be sum to $1$.






            share|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              You are right in a sense that it is better to be called log of unnormalized probability. This way, the quantity could be positive or negative. For example, $text{log}(0.5) < 0$ and $text{log}(12) > 0$ are both valid log of unnormalized probabilities. Here, in more detail:




              1. Probability: $P(i) = e^{o_i}/sum_{k=1}^{K}e^{o_k}$ (using softmax as mentioned in Figure 10.3 caption, and assuming $mathbf{o}=(o_1,..,o_K)$ is the output of layer before softmax),


              2. Unnormalized probability: $tilde{P}(i) = e^{o_i}$, which can be larger than 1,


              3. Log of unnormalized probability: $text{log}tilde{P}(i) = o_i$, which can be positive or negative.







              share|improve this answer











              $endgroup$


















                3












                $begingroup$

                You are right in a sense that it is better to be called log of unnormalized probability. This way, the quantity could be positive or negative. For example, $text{log}(0.5) < 0$ and $text{log}(12) > 0$ are both valid log of unnormalized probabilities. Here, in more detail:




                1. Probability: $P(i) = e^{o_i}/sum_{k=1}^{K}e^{o_k}$ (using softmax as mentioned in Figure 10.3 caption, and assuming $mathbf{o}=(o_1,..,o_K)$ is the output of layer before softmax),


                2. Unnormalized probability: $tilde{P}(i) = e^{o_i}$, which can be larger than 1,


                3. Log of unnormalized probability: $text{log}tilde{P}(i) = o_i$, which can be positive or negative.







                share|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You are right in a sense that it is better to be called log of unnormalized probability. This way, the quantity could be positive or negative. For example, $text{log}(0.5) < 0$ and $text{log}(12) > 0$ are both valid log of unnormalized probabilities. Here, in more detail:




                  1. Probability: $P(i) = e^{o_i}/sum_{k=1}^{K}e^{o_k}$ (using softmax as mentioned in Figure 10.3 caption, and assuming $mathbf{o}=(o_1,..,o_K)$ is the output of layer before softmax),


                  2. Unnormalized probability: $tilde{P}(i) = e^{o_i}$, which can be larger than 1,


                  3. Log of unnormalized probability: $text{log}tilde{P}(i) = o_i$, which can be positive or negative.







                  share|improve this answer











                  $endgroup$



                  You are right in a sense that it is better to be called log of unnormalized probability. This way, the quantity could be positive or negative. For example, $text{log}(0.5) < 0$ and $text{log}(12) > 0$ are both valid log of unnormalized probabilities. Here, in more detail:




                  1. Probability: $P(i) = e^{o_i}/sum_{k=1}^{K}e^{o_k}$ (using softmax as mentioned in Figure 10.3 caption, and assuming $mathbf{o}=(o_1,..,o_K)$ is the output of layer before softmax),


                  2. Unnormalized probability: $tilde{P}(i) = e^{o_i}$, which can be larger than 1,


                  3. Log of unnormalized probability: $text{log}tilde{P}(i) = o_i$, which can be positive or negative.








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited yesterday

























                  answered yesterday









                  EsmailianEsmailian

                  1,421113




                  1,421113























                      2












                      $begingroup$

                      You are right, nothing stop $o_k^{(t)}$ from being nonnegative, the keyword here is "unnormalized".



                      If we let $o_k^{(t)}=ln q_k^{(t)}$



                      $$hat{y}^{(t)}_k= frac{exp(o^{(t)}_k)}{sum_{k=1}^K exp(o^{(t)}_k) }= frac{q_k^{(t)}}{sum_{k=1}^K q_k^{(t)}}$$



                      Here $q_k^{(t)}$ can be any positive number, they will be normalized to be sum to $1$.






                      share|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        You are right, nothing stop $o_k^{(t)}$ from being nonnegative, the keyword here is "unnormalized".



                        If we let $o_k^{(t)}=ln q_k^{(t)}$



                        $$hat{y}^{(t)}_k= frac{exp(o^{(t)}_k)}{sum_{k=1}^K exp(o^{(t)}_k) }= frac{q_k^{(t)}}{sum_{k=1}^K q_k^{(t)}}$$



                        Here $q_k^{(t)}$ can be any positive number, they will be normalized to be sum to $1$.






                        share|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          You are right, nothing stop $o_k^{(t)}$ from being nonnegative, the keyword here is "unnormalized".



                          If we let $o_k^{(t)}=ln q_k^{(t)}$



                          $$hat{y}^{(t)}_k= frac{exp(o^{(t)}_k)}{sum_{k=1}^K exp(o^{(t)}_k) }= frac{q_k^{(t)}}{sum_{k=1}^K q_k^{(t)}}$$



                          Here $q_k^{(t)}$ can be any positive number, they will be normalized to be sum to $1$.






                          share|improve this answer









                          $endgroup$



                          You are right, nothing stop $o_k^{(t)}$ from being nonnegative, the keyword here is "unnormalized".



                          If we let $o_k^{(t)}=ln q_k^{(t)}$



                          $$hat{y}^{(t)}_k= frac{exp(o^{(t)}_k)}{sum_{k=1}^K exp(o^{(t)}_k) }= frac{q_k^{(t)}}{sum_{k=1}^K q_k^{(t)}}$$



                          Here $q_k^{(t)}$ can be any positive number, they will be normalized to be sum to $1$.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered yesterday









                          Siong Thye GohSiong Thye Goh

                          1,322418




                          1,322418






























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