Distribution of X-Y when X and Y are independent geometric
$X$ 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 Nov 19 '18 at 4:51
SolidSnake
102
add a comment |
$X$ 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 Nov 19 '18 at 4:51
SolidSnake
102
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
Nov 19 '18 at 9:00
add a comment |
$X$ 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 Nov 19 '18 at 4:51
SolidSnake
102
$X$ 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 Nov 19 '18 at 4:51
SolidSnake
102
asked Nov 19 '18 at 4:51
SolidSnake
102
asked Nov 19 '18 at 4:51
SolidSnake
102
asked Nov 19 '18 at 4:51
SolidSnake
102
asked Nov 19 '18 at 4:51
SolidSnake
102
102
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
Nov 19 '18 at 9:00
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
Nov 19 '18 at 9:00
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
Nov 19 '18 at 9:00
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
Nov 19 '18 at 9:00
add a comment |
1 Answer
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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 Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
add a comment |
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 Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
add a comment |
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 Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
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 Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
answered Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
answered Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
answered Nov 19 '18 at 4:59
Siong Thye Goh
99.3k1464117
99.3k1464117
add a comment |
add a comment |
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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
Nov 19 '18 at 9:00