Deriving likelihood function of binomial distribution, confusion over exponents












2














This question focuses on a specific aspect of this one:
How to derive the likelihood function for binomial distribution for parameter estimation?



In my own derivation, I start with:
$$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$



Ignoring $mC_x$, the likelihood function is then given by:



$$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$



However, in the question I referenced, they have this instead:
$$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$



My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?






estimation maximum-likelihood binomial likelihood point-estimation







share|cite|improve this question





























    2














    This question focuses on a specific aspect of this one:
    How to derive the likelihood function for binomial distribution for parameter estimation?



    In my own derivation, I start with:
    $$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$



    Ignoring $mC_x$, the likelihood function is then given by:



    $$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$



    However, in the question I referenced, they have this instead:
    $$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$



    My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?






    estimation maximum-likelihood binomial likelihood point-estimation







    share|cite|improve this question



























      2












      2








      2







      This question focuses on a specific aspect of this one:
      How to derive the likelihood function for binomial distribution for parameter estimation?



      In my own derivation, I start with:
      $$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$



      Ignoring $mC_x$, the likelihood function is then given by:



      $$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$



      However, in the question I referenced, they have this instead:
      $$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$



      My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?






      estimation maximum-likelihood binomial likelihood point-estimation







      share|cite|improve this question















      This question focuses on a specific aspect of this one:
      How to derive the likelihood function for binomial distribution for parameter estimation?



      In my own derivation, I start with:
      $$f(xmid p) = mC_x~p^x(1-p)^{m-x}$$



      Ignoring $mC_x$, the likelihood function is then given by:



      $$L(p) = prod_{i=1}^np^{x_i}(1-p)^{m-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n m-x_i} = p^{x}(1-p)^{nm-x}$$



      However, in the question I referenced, they have this instead:
      $$prod_{i=1}^np^{x_i}(1-p)^{1-x_i} = p^{sum_1^n x_i}(1-p)^{sum_1^n1-x_i} = p^{x}(1-p)^{n-x}$$



      My question is, are both approaches correct? If so, why does the referenced question use $1$ in place of $m$ on the exponents?






      estimation maximum-likelihood binomial likelihood point-estimation




      estimation maximum-likelihood binomial likelihood point-estimation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 23 at 18:09









      Michael Hardy

      3,5451430




      3,5451430










      asked Dec 23 at 17:11









      HumptyDumpty

      1183




      1183






















          3 Answers
          3






          active

          oldest

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          4














          It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
          $$
          L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
          $$
          It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
          $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
          $$
          L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
          $$
          Therefore both are right, but they're answers to different questions.






          share|cite|improve this answer





























            2














            The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.



            Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.



            Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.






            share|cite|improve this answer























            • Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
              – ruakh
              Dec 24 at 4:07










            • @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
              – StatsStudent
              2 days ago












            • But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
              – ruakh
              2 days ago










            • No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
              – StatsStudent
              2 days ago










            • That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
              – ruakh
              2 days ago



















            1














            The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.






            share|cite|improve this answer





















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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4














              It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
              $$
              L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
              $$
              It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
              $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
              $$
              L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
              $$
              Therefore both are right, but they're answers to different questions.






              share|cite|improve this answer


























                4














                It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
                $$
                L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
                $$
                It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
                $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
                $$
                L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
                $$
                Therefore both are right, but they're answers to different questions.






                share|cite|improve this answer
























                  4












                  4








                  4






                  It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
                  $$
                  L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
                  $$
                  It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
                  $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
                  $$
                  L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
                  $$
                  Therefore both are right, but they're answers to different questions.






                  share|cite|improve this answer












                  It looks as if you intended $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Binomial}(m,p).$ Then you have
                  $$
                  L(p) propto prod_{i=1}^n p^{x_i} (1-p)^{m-x_i} = p^{sum_{i=1}^n x_i} (1-p)^{nm - sum_{i=1}^n x_i} = p^x (1-p)^{nm-x}.
                  $$
                  It appears that in the question that you "referenced" (I don't know what "referenced" means in this context, but that doesn't appear to matter.) one has
                  $X_1,ldots,X_n sim operatorname{i{.}i{.}d{.}} operatorname{Bernoulli}(p)$, so that $X_1+cdots+X_n sim operatorname{Binomial} (n,p).$ That leads to
                  $$
                  L(p) propto p^{sum_{i=1}^n x_i} (1-p)^{n - sum_{i=1}^n x_i} = p^x(1-p)^{n-x}.
                  $$
                  Therefore both are right, but they're answers to different questions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 at 18:17









                  Michael Hardy

                  3,5451430




                  3,5451430

























                      2














                      The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.



                      Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.



                      Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.






                      share|cite|improve this answer























                      • Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
                        – ruakh
                        Dec 24 at 4:07










                      • @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
                        – StatsStudent
                        2 days ago












                      • But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
                        – ruakh
                        2 days ago










                      • No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
                        – StatsStudent
                        2 days ago










                      • That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
                        – ruakh
                        2 days ago
















                      2














                      The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.



                      Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.



                      Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.






                      share|cite|improve this answer























                      • Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
                        – ruakh
                        Dec 24 at 4:07










                      • @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
                        – StatsStudent
                        2 days ago












                      • But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
                        – ruakh
                        2 days ago










                      • No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
                        – StatsStudent
                        2 days ago










                      • That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
                        – ruakh
                        2 days ago














                      2












                      2








                      2






                      The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.



                      Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.



                      Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.






                      share|cite|improve this answer














                      The question you are referencing is starting with a Bernoulli distribution. To be sure, $x$ only takes on 0 or 1 in that question. In your work, you are starting with a Binomial distribution. To be sure, your values of $x=0, 1, 2, ... m$.



                      Remember that the sum of $n$ independent $Bernoulli(p)$ variables is a $Binomial(n, p)$ distribution. This should account for the differences you are seeing.



                      Your derivation for the likelihood of a binomial is just fine, ignoring the $mCx$ term, but you shouldn't ignore it. You can treat it as ignorable for the purposes of calculating the likelihood function since the likelihood function is a function only of the parameter $p$ and $p$ does not show up in $mCx$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 23 at 17:52

























                      answered Dec 23 at 17:36









                      StatsStudent

                      4,56732041




                      4,56732041












                      • Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
                        – ruakh
                        Dec 24 at 4:07










                      • @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
                        – StatsStudent
                        2 days ago












                      • But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
                        – ruakh
                        2 days ago










                      • No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
                        – StatsStudent
                        2 days ago










                      • That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
                        – ruakh
                        2 days ago


















                      • Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
                        – ruakh
                        Dec 24 at 4:07










                      • @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
                        – StatsStudent
                        2 days ago












                      • But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
                        – ruakh
                        2 days ago










                      • No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
                        – StatsStudent
                        2 days ago










                      • That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
                        – ruakh
                        2 days ago
















                      Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
                      – ruakh
                      Dec 24 at 4:07




                      Are you using "To be sure, [...]" to mean "You can be sure of this because [...]"? (I ask because "To be sure, [...]" means something quite different in English, which makes your usage very confusing. If you didn't intend the idiomatic meaning, I recommend finding a different phrase to express what you meant.)
                      – ruakh
                      Dec 24 at 4:07












                      @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
                      – StatsStudent
                      2 days ago






                      @ruakh, indeed the intention was idiomatic. It is correctly used to mean "certainly, undoubtedly, admittedly." I will keep it.
                      – StatsStudent
                      2 days ago














                      But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
                      – ruakh
                      2 days ago




                      But "admittedly" makes no sense in this context. You must be misunderstanding the idiom. :-/
                      – ruakh
                      2 days ago












                      No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
                      – StatsStudent
                      2 days ago




                      No, I have a perfect understanding. I'm using the idiomatic meaning of "certainly" or "undoubtedly" - not "admittedly."
                      – StatsStudent
                      2 days ago












                      That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
                      – ruakh
                      2 days ago




                      That's not a separate meaning. You can use "certainly" to mean "to be sure" (for example, "To be sure, not everyone does it; but most people do" can be rephrased as "Certainly, not everyone does it; but most people do"); but you can't use "to be sure" in all cases where you can use "certainly" (for example, "It's certainly possible; would you like us to do it now?" can't be rephrased as "To be sure, it's possible; would you like us to do it now?"). If the word "admittedly" doesn't work, then the phrase "to be sure" doesn't work, either.
                      – ruakh
                      2 days ago











                      1














                      The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.






                      share|cite|improve this answer


























                        1














                        The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.






                          share|cite|improve this answer












                          The PMF in the question, $f(x_i)=p^{x_i}(1-p)^{1-x_i}$ belongs to Bernoulli distribution, where $x_i$ is a binary variable. Yours is the PMF of Binomial distribution, and $x_i$'s are Binomial RVs ($n$ of them actually) with parameters $(m,p)$.







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                          answered Dec 23 at 17:39









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