Mixing
Distribution of X-Y when X and Y are independent geometric
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$X$ and $Y$ are independent geometric distributions with parameter $p$. Find the distribution for $X-Y$
I am aware that there are many similar questions. My problem with this specific transformation is that I cannot solve the summation I end up with.
This is what I have done:
$$Z = X - Y$$
$$P(Z=z) = P(X-Y=z)$$
$$= sum_{k=1}^infty P(Y=k)P(X=z+k)$$
$$= sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1}$$
Somehow the answer should be:
$$frac{p(1-p)^{|z|}}{2-p} $$
How does this sum turn into this final solution? Or have I done a mistake in the previous steps. Any help is appreciated, thanks!
probability probability-distributions summation transformation
asked 23 hours ago
SolidSnake
31
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up vote
0
down vote
favorite
$X$ and $Y$ are independent geometric distributions with parameter $p$. Find the distribution for $X-Y$
I am aware that there are many similar questions. My problem with this specific transformation is that I cannot solve the summation I end up with.
This is what I have done:
$$Z = X - Y$$
$$P(Z=z) = P(X-Y=z)$$
$$= sum_{k=1}^infty P(Y=k)P(X=z+k)$$
$$= sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1}$$
Somehow the answer should be:
$$frac{p(1-p)^{|z|}}{2-p} $$
How does this sum turn into this final solution? Or have I done a mistake in the previous steps. Any help is appreciated, thanks!
probability probability-distributions summation transformation
asked 23 hours ago
SolidSnake
31
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $z$ belongs to the set of all integers. There are two cases to consider when finding the distribution of $Z$: when $kge 1$ ( so that $zge 0$ ) and when $kge 1-z$ ( so that $z<0$ ). You have written the sum for the first case. Combining both cases you would get the final answer as given.
– StubbornAtom
19 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$X$ and $Y$ are independent geometric distributions with parameter $p$. Find the distribution for $X-Y$
I am aware that there are many similar questions. My problem with this specific transformation is that I cannot solve the summation I end up with.
This is what I have done:
$$Z = X - Y$$
$$P(Z=z) = P(X-Y=z)$$
$$= sum_{k=1}^infty P(Y=k)P(X=z+k)$$
$$= sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1}$$
Somehow the answer should be:
$$frac{p(1-p)^{|z|}}{2-p} $$
How does this sum turn into this final solution? Or have I done a mistake in the previous steps. Any help is appreciated, thanks!
probability probability-distributions summation transformation
asked 23 hours ago
SolidSnake
31
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$X$ and $Y$ are independent geometric distributions with parameter $p$. Find the distribution for $X-Y$
I am aware that there are many similar questions. My problem with this specific transformation is that I cannot solve the summation I end up with.
This is what I have done:
$$Z = X - Y$$
$$P(Z=z) = P(X-Y=z)$$
$$= sum_{k=1}^infty P(Y=k)P(X=z+k)$$
$$= sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1}$$
Somehow the answer should be:
$$frac{p(1-p)^{|z|}}{2-p} $$
How does this sum turn into this final solution? Or have I done a mistake in the previous steps. Any help is appreciated, thanks!
probability probability-distributions summation transformation
probability probability-distributions summation transformation
asked 23 hours ago
SolidSnake
31
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 hours ago
SolidSnake
31
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 hours ago
SolidSnake
31
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 hours ago
SolidSnake
31
asked 23 hours ago
SolidSnake
31
31
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
SolidSnake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $z$ belongs to the set of all integers. There are two cases to consider when finding the distribution of $Z$: when $kge 1$ ( so that $zge 0$ ) and when $kge 1-z$ ( so that $z<0$ ). You have written the sum for the first case. Combining both cases you would get the final answer as given.
– StubbornAtom
19 hours ago
add a comment |
Note that $z$ belongs to the set of all integers. There are two cases to consider when finding the distribution of $Z$: when $kge 1$ ( so that $zge 0$ ) and when $kge 1-z$ ( so that $z<0$ ). You have written the sum for the first case. Combining both cases you would get the final answer as given.
– StubbornAtom
19 hours ago
Note that $z$ belongs to the set of all integers. There are two cases to consider when finding the distribution of $Z$: when $kge 1$ ( so that $zge 0$ ) and when $kge 1-z$ ( so that $z<0$ ). You have written the sum for the first case. Combining both cases you would get the final answer as given.
– StubbornAtom
19 hours ago
Note that $z$ belongs to the set of all integers. There are two cases to consider when finding the distribution of $Z$: when $kge 1$ ( so that $zge 0$ ) and when $kge 1-z$ ( so that $z<0$ ). You have written the sum for the first case. Combining both cases you would get the final answer as given.
– StubbornAtom
19 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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accepted
Suppose $z ge 0$,begin{align}sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1} &= p^2sum_{k=1}^infty (1-p)^{z+2k-2} \
&=p^2(1-p)^z sum_{k=1}^infty ((1-p)^2)^{k-1}\
&=frac{p^2(1-p)^z}{1-(1-p)^2}\
&=frac{p^2(1-p)^z}{(2-p)p}\
&=frac{p(1-p)^z}{2-p}end{align}
answered 23 hours ago
Siong Thye Goh
92.9k1462114
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Suppose $z ge 0$,begin{align}sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1} &= p^2sum_{k=1}^infty (1-p)^{z+2k-2} \
&=p^2(1-p)^z sum_{k=1}^infty ((1-p)^2)^{k-1}\
&=frac{p^2(1-p)^z}{1-(1-p)^2}\
&=frac{p^2(1-p)^z}{(2-p)p}\
&=frac{p(1-p)^z}{2-p}end{align}
answered 23 hours ago
Siong Thye Goh
92.9k1462114
add a comment |
up vote
0
down vote
accepted
Suppose $z ge 0$,begin{align}sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1} &= p^2sum_{k=1}^infty (1-p)^{z+2k-2} \
&=p^2(1-p)^z sum_{k=1}^infty ((1-p)^2)^{k-1}\
&=frac{p^2(1-p)^z}{1-(1-p)^2}\
&=frac{p^2(1-p)^z}{(2-p)p}\
&=frac{p(1-p)^z}{2-p}end{align}
answered 23 hours ago
Siong Thye Goh
92.9k1462114
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Suppose $z ge 0$,begin{align}sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1} &= p^2sum_{k=1}^infty (1-p)^{z+2k-2} \
&=p^2(1-p)^z sum_{k=1}^infty ((1-p)^2)^{k-1}\
&=frac{p^2(1-p)^z}{1-(1-p)^2}\
&=frac{p^2(1-p)^z}{(2-p)p}\
&=frac{p(1-p)^z}{2-p}end{align}
answered 23 hours ago
Siong Thye Goh
92.9k1462114
Suppose $z ge 0$,begin{align}sum_{k=1}^infty p(1-p)^{k-1} p(1-p)^{z+k-1} &= p^2sum_{k=1}^infty (1-p)^{z+2k-2} \
&=p^2(1-p)^z sum_{k=1}^infty ((1-p)^2)^{k-1}\
&=frac{p^2(1-p)^z}{1-(1-p)^2}\
&=frac{p^2(1-p)^z}{(2-p)p}\
&=frac{p(1-p)^z}{2-p}end{align}
answered 23 hours ago
Siong Thye Goh
92.9k1462114
answered 23 hours ago
Siong Thye Goh
92.9k1462114
answered 23 hours ago
Siong Thye Goh
92.9k1462114
answered 23 hours ago
Siong Thye Goh
92.9k1462114
92.9k1462114
add a comment |
add a comment |
SolidSnake is a new contributor. Be nice, and check out our Code of Conduct.
SolidSnake is a new contributor. Be nice, and check out our Code of Conduct.
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Note that $z$ belongs to the set of all integers. There are two cases to consider when finding the distribution of $Z$: when $kge 1$ ( so that $zge 0$ ) and when $kge 1-z$ ( so that $z<0$ ). You have written the sum for the first case. Combining both cases you would get the final answer as given.
– StubbornAtom
19 hours ago