Quadratic expansion of function with random variable, deriving random variable

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Let $mathbf{x} in mathbb{R}^M$, $b in mathbb{R}$, $q_m=b cdot e^{-x_m}$, $p_m sim Poisson(b cdot e^{-x_m})$ be a Poisson random variable and



$$E(mathbf{x})=sum_{m=1}^M (b cdot e^{-x_m}-p_m cdot log(b cdot e^{x_m}))$$



In the paper I'm reading, it says, that the quadratic expansion with respect to $mathbf{x}$ around the point $-(ln(frac{q_m}{b}))$ is



$$Q(E(mathbf{mathbf{x}}))=sum_{m=1}^M frac{1}{2} bcdot e^{-x_m}(mathbf{x}-log(frac{b}{p_m}))^2$$




But how do I get there? (First of all, I think we could simplify $-(ln(frac{q_m}{b}))=x_m$, but that won't really help, so I'll just go with $-(ln(frac{q_m}{b}))$.)
As for the quadratic expansion, the formula is:



$$Q(E)= E(mathbf{x})+frac{d}{d mathbf{x}} E(mathbf{x}) + frac{1}{2} frac{d^2}{d mathbf{x}^2} E(mathbf{x}) \$$
$$=sum_{m=1}^M E(x_m)
+frac{d}{d mathbf{x}} left( sum_{m=1}^M E(x_m)right) (mathbf{x}+ln(frac{q_m}{b})) \
+frac{1}{2} frac{d^2}{d mathbf{x}^2} left( sum_{m=1}^M E(x_m) right) (mathbf{x}+ln(frac{q_m}{b}))^2$$



But how do I go on from here? Especially, how do I treat the random variable $p_m$ when deriving $E$?
Can I assume $p_m=q_m$? I did so and got:



$$E(x_m)=sum_{m=1}^M(bcdot e^{-x_m}-bcdot e^{-x_m} log(bcdot e^{-x_m}))$$
and
$$frac{d}{d mathbf{x}} sum_{m=1}^M E(x_m)= sum_{m=1}^M bcdot e^{-x_m} log(bcdot e^{-x_m})$$
and
$$frac{d^2}{d mathbf{x}^2} sum_{m=1}^M E(x_m)= sum_{m=1}^M b cdot e^{-x_m} (log(bcdot e^{-x_m})+1)$$



(I derived each summand by the $m$th component, is this correct?)
However, putting everything together, did not give me the expression from above, so I obviously made mistakes.



I'd be thankful for any kind of help. Thank you very much. (I also posted a more detailed version of this question some days ago here, but didn't get any responses, so I decided to make the question simpler. Of course, I will delete this first question, but I'll let it stay for the time being, in case someone's interested.)






taylor-expansion







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    $begingroup$


    Let $mathbf{x} in mathbb{R}^M$, $b in mathbb{R}$, $q_m=b cdot e^{-x_m}$, $p_m sim Poisson(b cdot e^{-x_m})$ be a Poisson random variable and



    $$E(mathbf{x})=sum_{m=1}^M (b cdot e^{-x_m}-p_m cdot log(b cdot e^{x_m}))$$



    In the paper I'm reading, it says, that the quadratic expansion with respect to $mathbf{x}$ around the point $-(ln(frac{q_m}{b}))$ is



    $$Q(E(mathbf{mathbf{x}}))=sum_{m=1}^M frac{1}{2} bcdot e^{-x_m}(mathbf{x}-log(frac{b}{p_m}))^2$$




    But how do I get there? (First of all, I think we could simplify $-(ln(frac{q_m}{b}))=x_m$, but that won't really help, so I'll just go with $-(ln(frac{q_m}{b}))$.)
    As for the quadratic expansion, the formula is:



    $$Q(E)= E(mathbf{x})+frac{d}{d mathbf{x}} E(mathbf{x}) + frac{1}{2} frac{d^2}{d mathbf{x}^2} E(mathbf{x}) \$$
    $$=sum_{m=1}^M E(x_m)
    +frac{d}{d mathbf{x}} left( sum_{m=1}^M E(x_m)right) (mathbf{x}+ln(frac{q_m}{b})) \
    +frac{1}{2} frac{d^2}{d mathbf{x}^2} left( sum_{m=1}^M E(x_m) right) (mathbf{x}+ln(frac{q_m}{b}))^2$$



    But how do I go on from here? Especially, how do I treat the random variable $p_m$ when deriving $E$?
    Can I assume $p_m=q_m$? I did so and got:



    $$E(x_m)=sum_{m=1}^M(bcdot e^{-x_m}-bcdot e^{-x_m} log(bcdot e^{-x_m}))$$
    and
    $$frac{d}{d mathbf{x}} sum_{m=1}^M E(x_m)= sum_{m=1}^M bcdot e^{-x_m} log(bcdot e^{-x_m})$$
    and
    $$frac{d^2}{d mathbf{x}^2} sum_{m=1}^M E(x_m)= sum_{m=1}^M b cdot e^{-x_m} (log(bcdot e^{-x_m})+1)$$



    (I derived each summand by the $m$th component, is this correct?)
    However, putting everything together, did not give me the expression from above, so I obviously made mistakes.



    I'd be thankful for any kind of help. Thank you very much. (I also posted a more detailed version of this question some days ago here, but didn't get any responses, so I decided to make the question simpler. Of course, I will delete this first question, but I'll let it stay for the time being, in case someone's interested.)






    taylor-expansion







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $mathbf{x} in mathbb{R}^M$, $b in mathbb{R}$, $q_m=b cdot e^{-x_m}$, $p_m sim Poisson(b cdot e^{-x_m})$ be a Poisson random variable and



      $$E(mathbf{x})=sum_{m=1}^M (b cdot e^{-x_m}-p_m cdot log(b cdot e^{x_m}))$$



      In the paper I'm reading, it says, that the quadratic expansion with respect to $mathbf{x}$ around the point $-(ln(frac{q_m}{b}))$ is



      $$Q(E(mathbf{mathbf{x}}))=sum_{m=1}^M frac{1}{2} bcdot e^{-x_m}(mathbf{x}-log(frac{b}{p_m}))^2$$




      But how do I get there? (First of all, I think we could simplify $-(ln(frac{q_m}{b}))=x_m$, but that won't really help, so I'll just go with $-(ln(frac{q_m}{b}))$.)
      As for the quadratic expansion, the formula is:



      $$Q(E)= E(mathbf{x})+frac{d}{d mathbf{x}} E(mathbf{x}) + frac{1}{2} frac{d^2}{d mathbf{x}^2} E(mathbf{x}) \$$
      $$=sum_{m=1}^M E(x_m)
      +frac{d}{d mathbf{x}} left( sum_{m=1}^M E(x_m)right) (mathbf{x}+ln(frac{q_m}{b})) \
      +frac{1}{2} frac{d^2}{d mathbf{x}^2} left( sum_{m=1}^M E(x_m) right) (mathbf{x}+ln(frac{q_m}{b}))^2$$



      But how do I go on from here? Especially, how do I treat the random variable $p_m$ when deriving $E$?
      Can I assume $p_m=q_m$? I did so and got:



      $$E(x_m)=sum_{m=1}^M(bcdot e^{-x_m}-bcdot e^{-x_m} log(bcdot e^{-x_m}))$$
      and
      $$frac{d}{d mathbf{x}} sum_{m=1}^M E(x_m)= sum_{m=1}^M bcdot e^{-x_m} log(bcdot e^{-x_m})$$
      and
      $$frac{d^2}{d mathbf{x}^2} sum_{m=1}^M E(x_m)= sum_{m=1}^M b cdot e^{-x_m} (log(bcdot e^{-x_m})+1)$$



      (I derived each summand by the $m$th component, is this correct?)
      However, putting everything together, did not give me the expression from above, so I obviously made mistakes.



      I'd be thankful for any kind of help. Thank you very much. (I also posted a more detailed version of this question some days ago here, but didn't get any responses, so I decided to make the question simpler. Of course, I will delete this first question, but I'll let it stay for the time being, in case someone's interested.)






      taylor-expansion







      share|cite|improve this question









      $endgroup$




      Let $mathbf{x} in mathbb{R}^M$, $b in mathbb{R}$, $q_m=b cdot e^{-x_m}$, $p_m sim Poisson(b cdot e^{-x_m})$ be a Poisson random variable and



      $$E(mathbf{x})=sum_{m=1}^M (b cdot e^{-x_m}-p_m cdot log(b cdot e^{x_m}))$$



      In the paper I'm reading, it says, that the quadratic expansion with respect to $mathbf{x}$ around the point $-(ln(frac{q_m}{b}))$ is



      $$Q(E(mathbf{mathbf{x}}))=sum_{m=1}^M frac{1}{2} bcdot e^{-x_m}(mathbf{x}-log(frac{b}{p_m}))^2$$




      But how do I get there? (First of all, I think we could simplify $-(ln(frac{q_m}{b}))=x_m$, but that won't really help, so I'll just go with $-(ln(frac{q_m}{b}))$.)
      As for the quadratic expansion, the formula is:



      $$Q(E)= E(mathbf{x})+frac{d}{d mathbf{x}} E(mathbf{x}) + frac{1}{2} frac{d^2}{d mathbf{x}^2} E(mathbf{x}) \$$
      $$=sum_{m=1}^M E(x_m)
      +frac{d}{d mathbf{x}} left( sum_{m=1}^M E(x_m)right) (mathbf{x}+ln(frac{q_m}{b})) \
      +frac{1}{2} frac{d^2}{d mathbf{x}^2} left( sum_{m=1}^M E(x_m) right) (mathbf{x}+ln(frac{q_m}{b}))^2$$



      But how do I go on from here? Especially, how do I treat the random variable $p_m$ when deriving $E$?
      Can I assume $p_m=q_m$? I did so and got:



      $$E(x_m)=sum_{m=1}^M(bcdot e^{-x_m}-bcdot e^{-x_m} log(bcdot e^{-x_m}))$$
      and
      $$frac{d}{d mathbf{x}} sum_{m=1}^M E(x_m)= sum_{m=1}^M bcdot e^{-x_m} log(bcdot e^{-x_m})$$
      and
      $$frac{d^2}{d mathbf{x}^2} sum_{m=1}^M E(x_m)= sum_{m=1}^M b cdot e^{-x_m} (log(bcdot e^{-x_m})+1)$$



      (I derived each summand by the $m$th component, is this correct?)
      However, putting everything together, did not give me the expression from above, so I obviously made mistakes.



      I'd be thankful for any kind of help. Thank you very much. (I also posted a more detailed version of this question some days ago here, but didn't get any responses, so I decided to make the question simpler. Of course, I will delete this first question, but I'll let it stay for the time being, in case someone's interested.)






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      asked Jan 9 at 14:56









      M. EvansM. Evans

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