In Information Theory, what is the lower bound that minimizes the value of $2^{l_x}$












0














Here is the question from my notes:



Suppose we wish to find a decipherable code that minimizes the expected value of $2^{l_x}$ for a probability distribution $p(x)$. Establish the lower bound
$$Ebig(2^{l_x}big)geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{1}$$
Hint: Use Cauchy-Schwarz



I am not really sure what to do in here but here is what I did:
$E(l_x)=sum_x p(x)l_x$ hence I say that
$$E(2^{l_x})=sum_x p(x)2^{l_x} tag{2}$$
Hence
$$sum_x p(x)2^{l_x}geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{3}$$
I take roots of both side and expand the summations:
$$sqrt{p(1)}+...sqrt{p(n)}leqsqrt{p(1)2^{l_1}+...p(n)2^{l_n}}tag{4}$$
The Cauchy Schwarz which I am supposed to use is I think something along those lines:
$$|x_1y_1+...+x_ny_n|leqsqrt{x_1^2+...x_n^2}sqrt{y_1^2+...y_n^2} tag{5}$$
Any hints on what to do here would be greatly appreciated.






probability inequality summation information-theory expected-value







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  • 2




    What is $l_x$ here?
    – Clement C.
    Nov 26 at 23:22










  • Length of the code word
    – Kudera Sebastian
    Nov 27 at 10:57
















0














Here is the question from my notes:



Suppose we wish to find a decipherable code that minimizes the expected value of $2^{l_x}$ for a probability distribution $p(x)$. Establish the lower bound
$$Ebig(2^{l_x}big)geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{1}$$
Hint: Use Cauchy-Schwarz



I am not really sure what to do in here but here is what I did:
$E(l_x)=sum_x p(x)l_x$ hence I say that
$$E(2^{l_x})=sum_x p(x)2^{l_x} tag{2}$$
Hence
$$sum_x p(x)2^{l_x}geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{3}$$
I take roots of both side and expand the summations:
$$sqrt{p(1)}+...sqrt{p(n)}leqsqrt{p(1)2^{l_1}+...p(n)2^{l_n}}tag{4}$$
The Cauchy Schwarz which I am supposed to use is I think something along those lines:
$$|x_1y_1+...+x_ny_n|leqsqrt{x_1^2+...x_n^2}sqrt{y_1^2+...y_n^2} tag{5}$$
Any hints on what to do here would be greatly appreciated.






probability inequality summation information-theory expected-value







share|cite|improve this question




















  • 2




    What is $l_x$ here?
    – Clement C.
    Nov 26 at 23:22










  • Length of the code word
    – Kudera Sebastian
    Nov 27 at 10:57














0












0








0







Here is the question from my notes:



Suppose we wish to find a decipherable code that minimizes the expected value of $2^{l_x}$ for a probability distribution $p(x)$. Establish the lower bound
$$Ebig(2^{l_x}big)geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{1}$$
Hint: Use Cauchy-Schwarz



I am not really sure what to do in here but here is what I did:
$E(l_x)=sum_x p(x)l_x$ hence I say that
$$E(2^{l_x})=sum_x p(x)2^{l_x} tag{2}$$
Hence
$$sum_x p(x)2^{l_x}geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{3}$$
I take roots of both side and expand the summations:
$$sqrt{p(1)}+...sqrt{p(n)}leqsqrt{p(1)2^{l_1}+...p(n)2^{l_n}}tag{4}$$
The Cauchy Schwarz which I am supposed to use is I think something along those lines:
$$|x_1y_1+...+x_ny_n|leqsqrt{x_1^2+...x_n^2}sqrt{y_1^2+...y_n^2} tag{5}$$
Any hints on what to do here would be greatly appreciated.






probability inequality summation information-theory expected-value







share|cite|improve this question















Here is the question from my notes:



Suppose we wish to find a decipherable code that minimizes the expected value of $2^{l_x}$ for a probability distribution $p(x)$. Establish the lower bound
$$Ebig(2^{l_x}big)geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{1}$$
Hint: Use Cauchy-Schwarz



I am not really sure what to do in here but here is what I did:
$E(l_x)=sum_x p(x)l_x$ hence I say that
$$E(2^{l_x})=sum_x p(x)2^{l_x} tag{2}$$
Hence
$$sum_x p(x)2^{l_x}geq bigg(sum_x sqrt{p(x)}bigg)^2 tag{3}$$
I take roots of both side and expand the summations:
$$sqrt{p(1)}+...sqrt{p(n)}leqsqrt{p(1)2^{l_1}+...p(n)2^{l_n}}tag{4}$$
The Cauchy Schwarz which I am supposed to use is I think something along those lines:
$$|x_1y_1+...+x_ny_n|leqsqrt{x_1^2+...x_n^2}sqrt{y_1^2+...y_n^2} tag{5}$$
Any hints on what to do here would be greatly appreciated.






probability inequality summation information-theory expected-value




probability inequality summation information-theory expected-value






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share|cite|improve this question













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edited Nov 27 at 13:32









leonbloy

40.3k645107




40.3k645107










asked Nov 26 at 23:10









Kudera Sebastian

534216




534216








  • 2




    What is $l_x$ here?
    – Clement C.
    Nov 26 at 23:22










  • Length of the code word
    – Kudera Sebastian
    Nov 27 at 10:57














  • 2




    What is $l_x$ here?
    – Clement C.
    Nov 26 at 23:22










  • Length of the code word
    – Kudera Sebastian
    Nov 27 at 10:57








2




2




What is $l_x$ here?
– Clement C.
Nov 26 at 23:22




What is $l_x$ here?
– Clement C.
Nov 26 at 23:22












Length of the code word
– Kudera Sebastian
Nov 27 at 10:57




Length of the code word
– Kudera Sebastian
Nov 27 at 10:57










1 Answer
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oldest

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2














Notice that we are speaking here of "decipherable codes", so we must put that restriction in play somehow. Fortunately we have Kraft-McMillan inequality.



So we must have $sum_x 2^{-l_x}le 1$.



Let's call $a_x = 2^{-l_x}$ and $b_x = p_x 2^{l_x}=p_x a_x^{-1} $



Then we want to bound $E=sum b_x$ subject to $sum a_x le 1$



Now, for any non-negative $a_x,b_x$ , we can write Cauchy-Schwarz as
$$sum_x sqrt{a_x b_x} le sqrt{ sum_x a_x}sqrt{ sum_x b_x} tag{1}$$



In our case this gives



$$sum_x sqrt{p_x} le sqrt{E} sqrt{ sum_x a_x} le sqrt{E} tag{2}$$



which gives the desired bound.






share|cite|improve this answer





















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    Notice that we are speaking here of "decipherable codes", so we must put that restriction in play somehow. Fortunately we have Kraft-McMillan inequality.



    So we must have $sum_x 2^{-l_x}le 1$.



    Let's call $a_x = 2^{-l_x}$ and $b_x = p_x 2^{l_x}=p_x a_x^{-1} $



    Then we want to bound $E=sum b_x$ subject to $sum a_x le 1$



    Now, for any non-negative $a_x,b_x$ , we can write Cauchy-Schwarz as
    $$sum_x sqrt{a_x b_x} le sqrt{ sum_x a_x}sqrt{ sum_x b_x} tag{1}$$



    In our case this gives



    $$sum_x sqrt{p_x} le sqrt{E} sqrt{ sum_x a_x} le sqrt{E} tag{2}$$



    which gives the desired bound.






    share|cite|improve this answer


























      2














      Notice that we are speaking here of "decipherable codes", so we must put that restriction in play somehow. Fortunately we have Kraft-McMillan inequality.



      So we must have $sum_x 2^{-l_x}le 1$.



      Let's call $a_x = 2^{-l_x}$ and $b_x = p_x 2^{l_x}=p_x a_x^{-1} $



      Then we want to bound $E=sum b_x$ subject to $sum a_x le 1$



      Now, for any non-negative $a_x,b_x$ , we can write Cauchy-Schwarz as
      $$sum_x sqrt{a_x b_x} le sqrt{ sum_x a_x}sqrt{ sum_x b_x} tag{1}$$



      In our case this gives



      $$sum_x sqrt{p_x} le sqrt{E} sqrt{ sum_x a_x} le sqrt{E} tag{2}$$



      which gives the desired bound.






      share|cite|improve this answer
























        2












        2








        2






        Notice that we are speaking here of "decipherable codes", so we must put that restriction in play somehow. Fortunately we have Kraft-McMillan inequality.



        So we must have $sum_x 2^{-l_x}le 1$.



        Let's call $a_x = 2^{-l_x}$ and $b_x = p_x 2^{l_x}=p_x a_x^{-1} $



        Then we want to bound $E=sum b_x$ subject to $sum a_x le 1$



        Now, for any non-negative $a_x,b_x$ , we can write Cauchy-Schwarz as
        $$sum_x sqrt{a_x b_x} le sqrt{ sum_x a_x}sqrt{ sum_x b_x} tag{1}$$



        In our case this gives



        $$sum_x sqrt{p_x} le sqrt{E} sqrt{ sum_x a_x} le sqrt{E} tag{2}$$



        which gives the desired bound.






        share|cite|improve this answer












        Notice that we are speaking here of "decipherable codes", so we must put that restriction in play somehow. Fortunately we have Kraft-McMillan inequality.



        So we must have $sum_x 2^{-l_x}le 1$.



        Let's call $a_x = 2^{-l_x}$ and $b_x = p_x 2^{l_x}=p_x a_x^{-1} $



        Then we want to bound $E=sum b_x$ subject to $sum a_x le 1$



        Now, for any non-negative $a_x,b_x$ , we can write Cauchy-Schwarz as
        $$sum_x sqrt{a_x b_x} le sqrt{ sum_x a_x}sqrt{ sum_x b_x} tag{1}$$



        In our case this gives



        $$sum_x sqrt{p_x} le sqrt{E} sqrt{ sum_x a_x} le sqrt{E} tag{2}$$



        which gives the desired bound.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 13:45









        leonbloy

        40.3k645107




        40.3k645107






























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