Sum of two multinomial random variables












1












$begingroup$


I have two independent multinomial random variables $Y_1$ and $Y_2$. I have to find the distribution of



$$X=Y_1+Y_2$$



$$Y_1 - text{Multinomial}(n_1,(p_1,p_2...p_k))$$
$$Y_2 - text{Multinomial}(n_2,(p_1,p_2...p_k))$$



I tried using the convolution to calculate the distribution but got stuck after a while



$$P(x_1,x_2..x_k) = sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k}p_1^{y_1}p_2^{y_2}..p_k^{y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)..(x_k-y_k)}p_1^{x_1-y_1}p_2^{x_2-y_2}..p_k^{x_k-y_k}$$



such that $y_1+y_2+...+y_n = n_1$ and by similar reasoning we see that $x_1+x_2+...+x_n=n_1+n_2$



$$P(x_1,x_2..x_k) = p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)...(x_k-y_k)}$$



$$P(x_1,x_2..x_k) = (n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} frac{1}{y_1! y_2!..y_k!} cdotfrac{1}{(x_1-y_1)! (x_2-y_2)!...(x_k-y_k)!}$$



$$P(x_1,x_2..x_k) = frac{(n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}}{x_1! x_2!..x_k!}sum_{y_1,y_2..y_n} binom{x_1}{y_1}binom{x_2}{y_2}...binom{x_k}{y_k}$$



But after this I couldn't solve it. Please help






probability statistics probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are they are independent?
    $endgroup$
    – Henry
    Jan 17 at 8:33










  • $begingroup$
    Yeah. They are independent.
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:04










  • $begingroup$
    Then, as $Y_1$ is the sum of $n_1$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ which is $text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$
    $endgroup$
    – Henry
    Jan 17 at 10:19










  • $begingroup$
    How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 10:57










  • $begingroup$
    It may depend on your definition of $text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point
    $endgroup$
    – Henry
    Jan 17 at 11:03
















1












$begingroup$


I have two independent multinomial random variables $Y_1$ and $Y_2$. I have to find the distribution of



$$X=Y_1+Y_2$$



$$Y_1 - text{Multinomial}(n_1,(p_1,p_2...p_k))$$
$$Y_2 - text{Multinomial}(n_2,(p_1,p_2...p_k))$$



I tried using the convolution to calculate the distribution but got stuck after a while



$$P(x_1,x_2..x_k) = sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k}p_1^{y_1}p_2^{y_2}..p_k^{y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)..(x_k-y_k)}p_1^{x_1-y_1}p_2^{x_2-y_2}..p_k^{x_k-y_k}$$



such that $y_1+y_2+...+y_n = n_1$ and by similar reasoning we see that $x_1+x_2+...+x_n=n_1+n_2$



$$P(x_1,x_2..x_k) = p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)...(x_k-y_k)}$$



$$P(x_1,x_2..x_k) = (n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} frac{1}{y_1! y_2!..y_k!} cdotfrac{1}{(x_1-y_1)! (x_2-y_2)!...(x_k-y_k)!}$$



$$P(x_1,x_2..x_k) = frac{(n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}}{x_1! x_2!..x_k!}sum_{y_1,y_2..y_n} binom{x_1}{y_1}binom{x_2}{y_2}...binom{x_k}{y_k}$$



But after this I couldn't solve it. Please help






probability statistics probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are they are independent?
    $endgroup$
    – Henry
    Jan 17 at 8:33










  • $begingroup$
    Yeah. They are independent.
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:04










  • $begingroup$
    Then, as $Y_1$ is the sum of $n_1$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ which is $text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$
    $endgroup$
    – Henry
    Jan 17 at 10:19










  • $begingroup$
    How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 10:57










  • $begingroup$
    It may depend on your definition of $text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point
    $endgroup$
    – Henry
    Jan 17 at 11:03














1












1








1





$begingroup$


I have two independent multinomial random variables $Y_1$ and $Y_2$. I have to find the distribution of



$$X=Y_1+Y_2$$



$$Y_1 - text{Multinomial}(n_1,(p_1,p_2...p_k))$$
$$Y_2 - text{Multinomial}(n_2,(p_1,p_2...p_k))$$



I tried using the convolution to calculate the distribution but got stuck after a while



$$P(x_1,x_2..x_k) = sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k}p_1^{y_1}p_2^{y_2}..p_k^{y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)..(x_k-y_k)}p_1^{x_1-y_1}p_2^{x_2-y_2}..p_k^{x_k-y_k}$$



such that $y_1+y_2+...+y_n = n_1$ and by similar reasoning we see that $x_1+x_2+...+x_n=n_1+n_2$



$$P(x_1,x_2..x_k) = p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)...(x_k-y_k)}$$



$$P(x_1,x_2..x_k) = (n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} frac{1}{y_1! y_2!..y_k!} cdotfrac{1}{(x_1-y_1)! (x_2-y_2)!...(x_k-y_k)!}$$



$$P(x_1,x_2..x_k) = frac{(n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}}{x_1! x_2!..x_k!}sum_{y_1,y_2..y_n} binom{x_1}{y_1}binom{x_2}{y_2}...binom{x_k}{y_k}$$



But after this I couldn't solve it. Please help






probability statistics probability-distributions







share|cite|improve this question











$endgroup$




I have two independent multinomial random variables $Y_1$ and $Y_2$. I have to find the distribution of



$$X=Y_1+Y_2$$



$$Y_1 - text{Multinomial}(n_1,(p_1,p_2...p_k))$$
$$Y_2 - text{Multinomial}(n_2,(p_1,p_2...p_k))$$



I tried using the convolution to calculate the distribution but got stuck after a while



$$P(x_1,x_2..x_k) = sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k}p_1^{y_1}p_2^{y_2}..p_k^{y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)..(x_k-y_k)}p_1^{x_1-y_1}p_2^{x_2-y_2}..p_k^{x_k-y_k}$$



such that $y_1+y_2+...+y_n = n_1$ and by similar reasoning we see that $x_1+x_2+...+x_n=n_1+n_2$



$$P(x_1,x_2..x_k) = p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} binom{n_1}{y_1 y_2..y_k} binom{n_2}{(x_1-y_1) (x_2-y_2)...(x_k-y_k)}$$



$$P(x_1,x_2..x_k) = (n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}sum_{y_1,y_2..y_n} frac{1}{y_1! y_2!..y_k!} cdotfrac{1}{(x_1-y_1)! (x_2-y_2)!...(x_k-y_k)!}$$



$$P(x_1,x_2..x_k) = frac{(n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}}{x_1! x_2!..x_k!}sum_{y_1,y_2..y_n} binom{x_1}{y_1}binom{x_2}{y_2}...binom{x_k}{y_k}$$



But after this I couldn't solve it. Please help






probability statistics probability-distributions




probability statistics probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 9:04







Sauhard Sharma

















asked Jan 17 at 5:49









Sauhard SharmaSauhard Sharma

953318




953318












  • $begingroup$
    Are they are independent?
    $endgroup$
    – Henry
    Jan 17 at 8:33










  • $begingroup$
    Yeah. They are independent.
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:04










  • $begingroup$
    Then, as $Y_1$ is the sum of $n_1$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ which is $text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$
    $endgroup$
    – Henry
    Jan 17 at 10:19










  • $begingroup$
    How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 10:57










  • $begingroup$
    It may depend on your definition of $text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point
    $endgroup$
    – Henry
    Jan 17 at 11:03


















  • $begingroup$
    Are they are independent?
    $endgroup$
    – Henry
    Jan 17 at 8:33










  • $begingroup$
    Yeah. They are independent.
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:04










  • $begingroup$
    Then, as $Y_1$ is the sum of $n_1$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ which is $text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$
    $endgroup$
    – Henry
    Jan 17 at 10:19










  • $begingroup$
    How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 10:57










  • $begingroup$
    It may depend on your definition of $text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point
    $endgroup$
    – Henry
    Jan 17 at 11:03
















$begingroup$
Are they are independent?
$endgroup$
– Henry
Jan 17 at 8:33




$begingroup$
Are they are independent?
$endgroup$
– Henry
Jan 17 at 8:33












$begingroup$
Yeah. They are independent.
$endgroup$
– Sauhard Sharma
Jan 17 at 9:04




$begingroup$
Yeah. They are independent.
$endgroup$
– Sauhard Sharma
Jan 17 at 9:04












$begingroup$
Then, as $Y_1$ is the sum of $n_1$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ which is $text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$
$endgroup$
– Henry
Jan 17 at 10:19




$begingroup$
Then, as $Y_1$ is the sum of $n_1$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $text{Multinomial}(1,(p_1,p_2...p_k))$ which is $text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$
$endgroup$
– Henry
Jan 17 at 10:19












$begingroup$
How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ?
$endgroup$
– Sauhard Sharma
Jan 17 at 10:57




$begingroup$
How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ?
$endgroup$
– Sauhard Sharma
Jan 17 at 10:57












$begingroup$
It may depend on your definition of $text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point
$endgroup$
– Henry
Jan 17 at 11:03




$begingroup$
It may depend on your definition of $text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point
$endgroup$
– Henry
Jan 17 at 11:03










1 Answer
1






active

oldest

votes


















1












$begingroup$

It would be easier to use characteristic functions.
begin{equation}
CF_{text{Multinomial}(n,(p_1,...,p_k))}(t_1,...,t_k) = bigg(sum_{j=1}^k p_je^{it_j}bigg)^n
end{equation}



As the CF of a sum of random variables is a product of their CFs, it is easy to spot that
begin{equation}
X sim text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))
end{equation}
as the equality of CFs induces equality of distributions and
begin{equation}
CF_X = CF_{Y_1+Y_2} = CF_{Y_1}CF_{Y_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1}bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1 + n_2}= CF_{text{Multinomial}(n_1 + n_2,(p_1,...,p_k))}(t_1,...,t_k).
end{equation}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
    $endgroup$
    – J.G.
    Jan 17 at 7:45












  • $begingroup$
    @J.G. Could you please do that and show me ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:09






  • 1




    $begingroup$
    @J.G. What is bad about complex numbers? :-)
    $endgroup$
    – Math-fun
    Jan 17 at 9:21










  • $begingroup$
    @SauhardSharma Just replace $e^{it_j}$ with $t_j$.
    $endgroup$
    – J.G.
    Jan 17 at 12:16












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

It would be easier to use characteristic functions.
begin{equation}
CF_{text{Multinomial}(n,(p_1,...,p_k))}(t_1,...,t_k) = bigg(sum_{j=1}^k p_je^{it_j}bigg)^n
end{equation}



As the CF of a sum of random variables is a product of their CFs, it is easy to spot that
begin{equation}
X sim text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))
end{equation}
as the equality of CFs induces equality of distributions and
begin{equation}
CF_X = CF_{Y_1+Y_2} = CF_{Y_1}CF_{Y_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1}bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1 + n_2}= CF_{text{Multinomial}(n_1 + n_2,(p_1,...,p_k))}(t_1,...,t_k).
end{equation}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
    $endgroup$
    – J.G.
    Jan 17 at 7:45












  • $begingroup$
    @J.G. Could you please do that and show me ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:09






  • 1




    $begingroup$
    @J.G. What is bad about complex numbers? :-)
    $endgroup$
    – Math-fun
    Jan 17 at 9:21










  • $begingroup$
    @SauhardSharma Just replace $e^{it_j}$ with $t_j$.
    $endgroup$
    – J.G.
    Jan 17 at 12:16
















1












$begingroup$

It would be easier to use characteristic functions.
begin{equation}
CF_{text{Multinomial}(n,(p_1,...,p_k))}(t_1,...,t_k) = bigg(sum_{j=1}^k p_je^{it_j}bigg)^n
end{equation}



As the CF of a sum of random variables is a product of their CFs, it is easy to spot that
begin{equation}
X sim text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))
end{equation}
as the equality of CFs induces equality of distributions and
begin{equation}
CF_X = CF_{Y_1+Y_2} = CF_{Y_1}CF_{Y_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1}bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1 + n_2}= CF_{text{Multinomial}(n_1 + n_2,(p_1,...,p_k))}(t_1,...,t_k).
end{equation}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
    $endgroup$
    – J.G.
    Jan 17 at 7:45












  • $begingroup$
    @J.G. Could you please do that and show me ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:09






  • 1




    $begingroup$
    @J.G. What is bad about complex numbers? :-)
    $endgroup$
    – Math-fun
    Jan 17 at 9:21










  • $begingroup$
    @SauhardSharma Just replace $e^{it_j}$ with $t_j$.
    $endgroup$
    – J.G.
    Jan 17 at 12:16














1












1








1





$begingroup$

It would be easier to use characteristic functions.
begin{equation}
CF_{text{Multinomial}(n,(p_1,...,p_k))}(t_1,...,t_k) = bigg(sum_{j=1}^k p_je^{it_j}bigg)^n
end{equation}



As the CF of a sum of random variables is a product of their CFs, it is easy to spot that
begin{equation}
X sim text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))
end{equation}
as the equality of CFs induces equality of distributions and
begin{equation}
CF_X = CF_{Y_1+Y_2} = CF_{Y_1}CF_{Y_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1}bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1 + n_2}= CF_{text{Multinomial}(n_1 + n_2,(p_1,...,p_k))}(t_1,...,t_k).
end{equation}






share|cite|improve this answer











$endgroup$



It would be easier to use characteristic functions.
begin{equation}
CF_{text{Multinomial}(n,(p_1,...,p_k))}(t_1,...,t_k) = bigg(sum_{j=1}^k p_je^{it_j}bigg)^n
end{equation}



As the CF of a sum of random variables is a product of their CFs, it is easy to spot that
begin{equation}
X sim text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))
end{equation}
as the equality of CFs induces equality of distributions and
begin{equation}
CF_X = CF_{Y_1+Y_2} = CF_{Y_1}CF_{Y_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1}bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_2} = bigg(sum_{j=1}^k p_je^{it_j}bigg)^{n_1 + n_2}= CF_{text{Multinomial}(n_1 + n_2,(p_1,...,p_k))}(t_1,...,t_k).
end{equation}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 6:45

























answered Jan 17 at 7:07









vermatorvermator

36110




36110












  • $begingroup$
    Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
    $endgroup$
    – J.G.
    Jan 17 at 7:45












  • $begingroup$
    @J.G. Could you please do that and show me ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:09






  • 1




    $begingroup$
    @J.G. What is bad about complex numbers? :-)
    $endgroup$
    – Math-fun
    Jan 17 at 9:21










  • $begingroup$
    @SauhardSharma Just replace $e^{it_j}$ with $t_j$.
    $endgroup$
    – J.G.
    Jan 17 at 12:16


















  • $begingroup$
    Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
    $endgroup$
    – J.G.
    Jan 17 at 7:45












  • $begingroup$
    @J.G. Could you please do that and show me ?
    $endgroup$
    – Sauhard Sharma
    Jan 17 at 9:09






  • 1




    $begingroup$
    @J.G. What is bad about complex numbers? :-)
    $endgroup$
    – Math-fun
    Jan 17 at 9:21










  • $begingroup$
    @SauhardSharma Just replace $e^{it_j}$ with $t_j$.
    $endgroup$
    – J.G.
    Jan 17 at 12:16
















$begingroup$
Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
$endgroup$
– J.G.
Jan 17 at 7:45






$begingroup$
Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to.
$endgroup$
– J.G.
Jan 17 at 7:45














$begingroup$
@J.G. Could you please do that and show me ?
$endgroup$
– Sauhard Sharma
Jan 17 at 9:09




$begingroup$
@J.G. Could you please do that and show me ?
$endgroup$
– Sauhard Sharma
Jan 17 at 9:09




1




1




$begingroup$
@J.G. What is bad about complex numbers? :-)
$endgroup$
– Math-fun
Jan 17 at 9:21




$begingroup$
@J.G. What is bad about complex numbers? :-)
$endgroup$
– Math-fun
Jan 17 at 9:21












$begingroup$
@SauhardSharma Just replace $e^{it_j}$ with $t_j$.
$endgroup$
– J.G.
Jan 17 at 12:16




$begingroup$
@SauhardSharma Just replace $e^{it_j}$ with $t_j$.
$endgroup$
– J.G.
Jan 17 at 12:16


















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