A experiment of throwing two fair cubic dice

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Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?






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    $begingroup$


    Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?






    probability







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?






      probability







      share|cite|improve this question









      $endgroup$




      Two fair cubic dice are thrown repeatedly in an experiment. Let $X_i$ be the absolute difference of the value of the two dice in the $i$-th throw. The experiment will be stopped when $X_i=0$. Let $Y=sum X_i$. How can I evaluate the mean and variance of Y, if the experiment is stopped at the $n$-th throw? And if the experiment is stopped at the 4-th throw and $Y=5$, what is the probability of $X_1=1$?






      probability




      probability






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      asked Jan 22 at 8:09









      Weihao HuangWeihao Huang

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          $begingroup$

          Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.



          Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$



          For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$



          To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.



          Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.




          To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
            $endgroup$
            – Weihao Huang
            Jan 22 at 10:48










          • $begingroup$
            Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
            $endgroup$
            – drhab
            Jan 22 at 11:26












          • $begingroup$
            I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:33










          • $begingroup$
            "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
            $endgroup$
            – drhab
            Jan 22 at 11:38










          • $begingroup$
            I see what you mean, thanks
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:59











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          $begingroup$

          Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.



          Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$



          For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$



          To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.



          Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.




          To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
            $endgroup$
            – Weihao Huang
            Jan 22 at 10:48










          • $begingroup$
            Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
            $endgroup$
            – drhab
            Jan 22 at 11:26












          • $begingroup$
            I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:33










          • $begingroup$
            "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
            $endgroup$
            – drhab
            Jan 22 at 11:38










          • $begingroup$
            I see what you mean, thanks
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:59
















          0












          $begingroup$

          Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.



          Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$



          For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$



          To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.



          Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.




          To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
            $endgroup$
            – Weihao Huang
            Jan 22 at 10:48










          • $begingroup$
            Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
            $endgroup$
            – drhab
            Jan 22 at 11:26












          • $begingroup$
            I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:33










          • $begingroup$
            "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
            $endgroup$
            – drhab
            Jan 22 at 11:38










          • $begingroup$
            I see what you mean, thanks
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:59














          0












          0








          0





          $begingroup$

          Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.



          Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$



          For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$



          To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.



          Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.




          To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$






          share|cite|improve this answer









          $endgroup$



          Let $E_n$ denote the event that the experiment is stopped at the $n$-th throw.



          Then to be found is:$$mathbb E[Ymid E_n]=sum_{i=1}^{n-1}mathbb E[X_imid E_n]=(n-1)mathbb E[X_1mid E_n]=(n-1)mathbb E[X_1mid X_1>0]$$



          For the last equality observe that $E_n={X_1>0}cap F_n$ where $F_n:={X_2>0,cdots,X_{n-1}>0,X_n=0}$ is an event such that $mathbf1_{F_n}$ and $X_1$ are independent so that:$$mathbb E[X_1mid E_n]=frac{mathbb EX_1mathbf1_{X_1>0}mathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}mathbb Emathbf1_{F_n}}{mathbb Emathbf1_{X_1>0}mathbb Emathbf1_{F_n}}=frac{mathbb EX_1mathbf1_{X_1>0}}{mathbb Emathbf1_{X_1>0}}=mathbb E[X_1mid X_1>0]$$



          To find the mean of $Y$ it is enough now to find $mathbb E[X_1mid X_1>0]$ and I leave that to you.



          Similarly we can find: $$mathbb E[Y^2mid E_n]=$$$$sum_{i=1}^{n-1}sum_{j=1}^{n-1}mathbb E[X_iX_jmid E_n]=(n-1)(n-2)mathbb E[X_1X_2mid X_1>0,X_2>0]+(n-1)mathbb E[X_1^2mid X_1>0]$$and then variance $mathbb E[Y^2mid E_n]-(mathbb E[Ymid E_n])^2$.




          To be found is: $$P(X_1=1mid X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}{P(X_1>0,X_2>0,X_3>0,X_4=0,X_1+X_2+X_3=5)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)P(X_4=0)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)P(X_4=0)}=$$$$frac{P(X_1=1,X_2>0,X_3>0,X_2+X_3=4)}{P(X_1>0,X_2>0,X_3>0,X_1+X_2+X_3=5)}=$$$$frac{sum_{s,t>0,s+t=4}P(X_1=1)P(X_2=s)P(X_3=t)}{sum_{r,s,t>0,r+s+t=5}P(X_1=r)P(X_2=s)P(X_3=t)}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 22 at 9:58









          drhabdrhab

          102k545136




          102k545136












          • $begingroup$
            Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
            $endgroup$
            – Weihao Huang
            Jan 22 at 10:48










          • $begingroup$
            Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
            $endgroup$
            – drhab
            Jan 22 at 11:26












          • $begingroup$
            I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:33










          • $begingroup$
            "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
            $endgroup$
            – drhab
            Jan 22 at 11:38










          • $begingroup$
            I see what you mean, thanks
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:59


















          • $begingroup$
            Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
            $endgroup$
            – Weihao Huang
            Jan 22 at 10:48










          • $begingroup$
            Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
            $endgroup$
            – drhab
            Jan 22 at 11:26












          • $begingroup$
            I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:33










          • $begingroup$
            "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
            $endgroup$
            – drhab
            Jan 22 at 11:38










          • $begingroup$
            I see what you mean, thanks
            $endgroup$
            – Weihao Huang
            Jan 22 at 11:59
















          $begingroup$
          Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
          $endgroup$
          – Weihao Huang
          Jan 22 at 10:48




          $begingroup$
          Thanks for the reply. Since I not good at probability, I hope you can confirm my works. The mean is $E(Y)=(n-1)E(X_i)$ and the variance is $Var(Y)=(n-1)[E(X_i^2)-(E(X_i))^2]$. I don't understand why you write $X_1>0$ as a condition since the value of $X_i$ has already grater than 0.
          $endgroup$
          – Weihao Huang
          Jan 22 at 10:48












          $begingroup$
          Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
          $endgroup$
          – drhab
          Jan 22 at 11:26






          $begingroup$
          Do you mean that $X_i>0$ (where $i>1$) already implies that $X_1>0$? I do that because I look at this as a process that does not stop (that does not harm, $Y$ can still be defined properly), and work with $X_i$ that are independent. If e.g. conditions like $X_2>0implies X_1>0$ are build in (I avoid that) then this (valuable) independence is lost.
          $endgroup$
          – drhab
          Jan 22 at 11:26














          $begingroup$
          I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
          $endgroup$
          – Weihao Huang
          Jan 22 at 11:33




          $begingroup$
          I see, it not that complicated. $X_i$ is defined as the $|D_1-D_2|$, so it must be greater than $0$.
          $endgroup$
          – Weihao Huang
          Jan 22 at 11:33












          $begingroup$
          "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
          $endgroup$
          – drhab
          Jan 22 at 11:38




          $begingroup$
          "..so it must be greater than $0$." What do you mean? For every index $i$ it is possible that $D_1^{(i)}=D_2^{(i)}$ or equivalently $X_i=0$.
          $endgroup$
          – drhab
          Jan 22 at 11:38












          $begingroup$
          I see what you mean, thanks
          $endgroup$
          – Weihao Huang
          Jan 22 at 11:59




          $begingroup$
          I see what you mean, thanks
          $endgroup$
          – Weihao Huang
          Jan 22 at 11:59


















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