Mixing
Sum of two Geometric Random Variables with different success probability
$begingroup$
Here is a problem and my answer which I know is wrong. I am hoping somebody can tell me where I went wrong.
Thanks
Bob
Problem:
Let $X$ and $Y$ be independent random variables having geometric
densities with parameters $p_1$ and $p_2$ respectively. Find the density of
$X + Y$.
Answer:
Let $Z = X + Y$.
begin{eqnarray*}
P( Z = 0 ) &=& P( X = 0 ) P(Y = 0) \
P( Z = 1 ) &=& P( X = 0 ) P(Y = 1) + P( X = 1 ) P(Y = 0) \
P( Z = n ) &=& sum_{i = 0}^{n} P(X = i)P(Y = n - i ) = sum_{i = 0}^{n} p_1(1-p_1)^i p_2 (1 - p_2)^{n - i} \
end{eqnarray*}
This is a finite geometric series with $a = p_1 p_2$ and $r = frac{1-p_1}{1-p_2}$. Also observe that there are $n+1$ terms not $n$ terms.
begin{eqnarray*}
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}{1 - frac{1-p_1}{1-p_2}} \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
end{eqnarray*}
probability probability-distributions
edited Jan 10 at 2:56
amWhy
1
asked Feb 9 '18 at 21:47
BobBob
923515
$endgroup$
add a comment |
$begingroup$
Here is a problem and my answer which I know is wrong. I am hoping somebody can tell me where I went wrong.
Thanks
Bob
Problem:
Let $X$ and $Y$ be independent random variables having geometric
densities with parameters $p_1$ and $p_2$ respectively. Find the density of
$X + Y$.
Answer:
Let $Z = X + Y$.
begin{eqnarray*}
P( Z = 0 ) &=& P( X = 0 ) P(Y = 0) \
P( Z = 1 ) &=& P( X = 0 ) P(Y = 1) + P( X = 1 ) P(Y = 0) \
P( Z = n ) &=& sum_{i = 0}^{n} P(X = i)P(Y = n - i ) = sum_{i = 0}^{n} p_1(1-p_1)^i p_2 (1 - p_2)^{n - i} \
end{eqnarray*}
This is a finite geometric series with $a = p_1 p_2$ and $r = frac{1-p_1}{1-p_2}$. Also observe that there are $n+1$ terms not $n$ terms.
begin{eqnarray*}
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}{1 - frac{1-p_1}{1-p_2}} \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
end{eqnarray*}
probability probability-distributions
edited Jan 10 at 2:56
amWhy
1
asked Feb 9 '18 at 21:47
BobBob
923515
$endgroup$
2
$begingroup$
Tiny mistake - $a$ is $p_1 p_2 (1-p_2)^{n},$ not $p_1p_2$.
$endgroup$
– stochasticboy321
Feb 9 '18 at 22:14
$begingroup$
are you sure the question says density?
$endgroup$
– Henry
Feb 9 '18 at 22:54
$begingroup$
@Henry The book does say density but I believe the modern term probability mass function.
$endgroup$
– Bob
Feb 9 '18 at 23:03
add a comment |
$begingroup$
Here is a problem and my answer which I know is wrong. I am hoping somebody can tell me where I went wrong.
Thanks
Bob
Problem:
Let $X$ and $Y$ be independent random variables having geometric
densities with parameters $p_1$ and $p_2$ respectively. Find the density of
$X + Y$.
Answer:
Let $Z = X + Y$.
begin{eqnarray*}
P( Z = 0 ) &=& P( X = 0 ) P(Y = 0) \
P( Z = 1 ) &=& P( X = 0 ) P(Y = 1) + P( X = 1 ) P(Y = 0) \
P( Z = n ) &=& sum_{i = 0}^{n} P(X = i)P(Y = n - i ) = sum_{i = 0}^{n} p_1(1-p_1)^i p_2 (1 - p_2)^{n - i} \
end{eqnarray*}
This is a finite geometric series with $a = p_1 p_2$ and $r = frac{1-p_1}{1-p_2}$. Also observe that there are $n+1$ terms not $n$ terms.
begin{eqnarray*}
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}{1 - frac{1-p_1}{1-p_2}} \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
end{eqnarray*}
probability probability-distributions
edited Jan 10 at 2:56
amWhy
1
asked Feb 9 '18 at 21:47
BobBob
923515
$endgroup$
Here is a problem and my answer which I know is wrong. I am hoping somebody can tell me where I went wrong.
Thanks
Bob
Problem:
Let $X$ and $Y$ be independent random variables having geometric
densities with parameters $p_1$ and $p_2$ respectively. Find the density of
$X + Y$.
Answer:
Let $Z = X + Y$.
begin{eqnarray*}
P( Z = 0 ) &=& P( X = 0 ) P(Y = 0) \
P( Z = 1 ) &=& P( X = 0 ) P(Y = 1) + P( X = 1 ) P(Y = 0) \
P( Z = n ) &=& sum_{i = 0}^{n} P(X = i)P(Y = n - i ) = sum_{i = 0}^{n} p_1(1-p_1)^i p_2 (1 - p_2)^{n - i} \
end{eqnarray*}
This is a finite geometric series with $a = p_1 p_2$ and $r = frac{1-p_1}{1-p_2}$. Also observe that there are $n+1$ terms not $n$ terms.
begin{eqnarray*}
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}{1 - frac{1-p_1}{1-p_2}} \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
P( Z = n ) &=&
frac{p_1 p_2Big(1 - {Big(frac{1-p_1}{1-p_2}Big) } ^{n+1}Big)}
{ frac{p_1 - p_2}{1 - p_2} } \
end{eqnarray*}
probability probability-distributions
probability probability-distributions
edited Jan 10 at 2:56
amWhy
1
asked Feb 9 '18 at 21:47
BobBob
923515
edited Jan 10 at 2:56
amWhy
1
asked Feb 9 '18 at 21:47
BobBob
923515
edited Jan 10 at 2:56
amWhy
1
edited Jan 10 at 2:56
amWhy
1
edited Jan 10 at 2:56
amWhy
1
1
asked Feb 9 '18 at 21:47
BobBob
923515
asked Feb 9 '18 at 21:47
BobBob
923515
asked Feb 9 '18 at 21:47
BobBob
923515
923515
2
$begingroup$
Tiny mistake - $a$ is $p_1 p_2 (1-p_2)^{n},$ not $p_1p_2$.
$endgroup$
– stochasticboy321
Feb 9 '18 at 22:14
$begingroup$
are you sure the question says density?
$endgroup$
– Henry
Feb 9 '18 at 22:54
$begingroup$
@Henry The book does say density but I believe the modern term probability mass function.
$endgroup$
– Bob
Feb 9 '18 at 23:03
add a comment |
2
$begingroup$
Tiny mistake - $a$ is $p_1 p_2 (1-p_2)^{n},$ not $p_1p_2$.
$endgroup$
– stochasticboy321
Feb 9 '18 at 22:14
$begingroup$
are you sure the question says density?
$endgroup$
– Henry
Feb 9 '18 at 22:54
$begingroup$
@Henry The book does say density but I believe the modern term probability mass function.
$endgroup$
– Bob
Feb 9 '18 at 23:03
2
2
$begingroup$
Tiny mistake - $a$ is $p_1 p_2 (1-p_2)^{n},$ not $p_1p_2$.
$endgroup$
– stochasticboy321
Feb 9 '18 at 22:14
$begingroup$
Tiny mistake - $a$ is $p_1 p_2 (1-p_2)^{n},$ not $p_1p_2$.
$endgroup$
– stochasticboy321
Feb 9 '18 at 22:14
$begingroup$
are you sure the question says density?
$endgroup$
– Henry
Feb 9 '18 at 22:54
$begingroup$
are you sure the question says density?
$endgroup$
– Henry
Feb 9 '18 at 22:54
$begingroup$
@Henry The book does say density but I believe the modern term probability mass function.
$endgroup$
– Bob
Feb 9 '18 at 23:03
$begingroup$
@Henry The book does say density but I believe the modern term probability mass function.
$endgroup$
– Bob
Feb 9 '18 at 23:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your definition of a geometric random variable is not quite consistent with the normal definition; normally one would say that $X$ is the trial on which one has the first success (in a sequence of $p_1$ Benoulli variables).
So that means
$$mathbf{P}(X = k) = (1-p_1)^{k-1} p_1, qquad k = 1,2,ldots$$
In particular the distribution is defined only for integers greater than or equal to $1$. In your definition (which I will denote $widehat X$, you allow $widehat X = 0$ to be non-zero; that is you assume the density is
$$mathbf{P}left( widehat X = kright) = (1-p_1)^k p_1, qquad k = 0,1,ldots$$
This also has an interpretation: this is you are counting the number of failures before success, so your definition is equivalent to
$$ widehat X = X-1.$$
From here we can determine the distribution of $Z = X + Y$ using the method you have
begin{align*}
mathbf{P}(Z = n) & = sum_{k=0}^n P(X = k) P(Y = n-k) \
& = sum_{k=1}^{n-1} P(X = k) P(Y = n-k) \
& = p_1 p_2 sum_{k=1}^{n-1} (1-p_1)^{k-1} (1-p_2)^{n-k-1} \
& = p_1 p_2frac{(1-p_2)^{n-1}}{(1-p_1)} sum_{k=1}^{n-1} left( frac{1-p_1}{1-p_2} right)^{k}.
end{align*}
From here you can manipulate the geometric series (much as you do above) to derive
$$mathbf{P}(Z = n) = frac{p_1p_2}{p_1 - p_2}
big( (1-p_2)^{n-1} - (1-p_1)^{n-1} big), qquad n = 2,3,ldots
$$
Note that $Z$ can only take values greater than or equal to $2$.
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Your definition of a geometric random variable is not quite consistent with the normal definition; normally one would say that $X$ is the trial on which one has the first success (in a sequence of $p_1$ Benoulli variables).
So that means
$$mathbf{P}(X = k) = (1-p_1)^{k-1} p_1, qquad k = 1,2,ldots$$
In particular the distribution is defined only for integers greater than or equal to $1$. In your definition (which I will denote $widehat X$, you allow $widehat X = 0$ to be non-zero; that is you assume the density is
$$mathbf{P}left( widehat X = kright) = (1-p_1)^k p_1, qquad k = 0,1,ldots$$
This also has an interpretation: this is you are counting the number of failures before success, so your definition is equivalent to
$$ widehat X = X-1.$$
From here we can determine the distribution of $Z = X + Y$ using the method you have
begin{align*}
mathbf{P}(Z = n) & = sum_{k=0}^n P(X = k) P(Y = n-k) \
& = sum_{k=1}^{n-1} P(X = k) P(Y = n-k) \
& = p_1 p_2 sum_{k=1}^{n-1} (1-p_1)^{k-1} (1-p_2)^{n-k-1} \
& = p_1 p_2frac{(1-p_2)^{n-1}}{(1-p_1)} sum_{k=1}^{n-1} left( frac{1-p_1}{1-p_2} right)^{k}.
end{align*}
From here you can manipulate the geometric series (much as you do above) to derive
$$mathbf{P}(Z = n) = frac{p_1p_2}{p_1 - p_2}
big( (1-p_2)^{n-1} - (1-p_1)^{n-1} big), qquad n = 2,3,ldots
$$
Note that $Z$ can only take values greater than or equal to $2$.
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
$endgroup$
add a comment |
$begingroup$
Your definition of a geometric random variable is not quite consistent with the normal definition; normally one would say that $X$ is the trial on which one has the first success (in a sequence of $p_1$ Benoulli variables).
So that means
$$mathbf{P}(X = k) = (1-p_1)^{k-1} p_1, qquad k = 1,2,ldots$$
In particular the distribution is defined only for integers greater than or equal to $1$. In your definition (which I will denote $widehat X$, you allow $widehat X = 0$ to be non-zero; that is you assume the density is
$$mathbf{P}left( widehat X = kright) = (1-p_1)^k p_1, qquad k = 0,1,ldots$$
This also has an interpretation: this is you are counting the number of failures before success, so your definition is equivalent to
$$ widehat X = X-1.$$
From here we can determine the distribution of $Z = X + Y$ using the method you have
begin{align*}
mathbf{P}(Z = n) & = sum_{k=0}^n P(X = k) P(Y = n-k) \
& = sum_{k=1}^{n-1} P(X = k) P(Y = n-k) \
& = p_1 p_2 sum_{k=1}^{n-1} (1-p_1)^{k-1} (1-p_2)^{n-k-1} \
& = p_1 p_2frac{(1-p_2)^{n-1}}{(1-p_1)} sum_{k=1}^{n-1} left( frac{1-p_1}{1-p_2} right)^{k}.
end{align*}
From here you can manipulate the geometric series (much as you do above) to derive
$$mathbf{P}(Z = n) = frac{p_1p_2}{p_1 - p_2}
big( (1-p_2)^{n-1} - (1-p_1)^{n-1} big), qquad n = 2,3,ldots
$$
Note that $Z$ can only take values greater than or equal to $2$.
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
$endgroup$
add a comment |
$begingroup$
Your definition of a geometric random variable is not quite consistent with the normal definition; normally one would say that $X$ is the trial on which one has the first success (in a sequence of $p_1$ Benoulli variables).
So that means
$$mathbf{P}(X = k) = (1-p_1)^{k-1} p_1, qquad k = 1,2,ldots$$
In particular the distribution is defined only for integers greater than or equal to $1$. In your definition (which I will denote $widehat X$, you allow $widehat X = 0$ to be non-zero; that is you assume the density is
$$mathbf{P}left( widehat X = kright) = (1-p_1)^k p_1, qquad k = 0,1,ldots$$
This also has an interpretation: this is you are counting the number of failures before success, so your definition is equivalent to
$$ widehat X = X-1.$$
From here we can determine the distribution of $Z = X + Y$ using the method you have
begin{align*}
mathbf{P}(Z = n) & = sum_{k=0}^n P(X = k) P(Y = n-k) \
& = sum_{k=1}^{n-1} P(X = k) P(Y = n-k) \
& = p_1 p_2 sum_{k=1}^{n-1} (1-p_1)^{k-1} (1-p_2)^{n-k-1} \
& = p_1 p_2frac{(1-p_2)^{n-1}}{(1-p_1)} sum_{k=1}^{n-1} left( frac{1-p_1}{1-p_2} right)^{k}.
end{align*}
From here you can manipulate the geometric series (much as you do above) to derive
$$mathbf{P}(Z = n) = frac{p_1p_2}{p_1 - p_2}
big( (1-p_2)^{n-1} - (1-p_1)^{n-1} big), qquad n = 2,3,ldots
$$
Note that $Z$ can only take values greater than or equal to $2$.
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
$endgroup$
Your definition of a geometric random variable is not quite consistent with the normal definition; normally one would say that $X$ is the trial on which one has the first success (in a sequence of $p_1$ Benoulli variables).
So that means
$$mathbf{P}(X = k) = (1-p_1)^{k-1} p_1, qquad k = 1,2,ldots$$
In particular the distribution is defined only for integers greater than or equal to $1$. In your definition (which I will denote $widehat X$, you allow $widehat X = 0$ to be non-zero; that is you assume the density is
$$mathbf{P}left( widehat X = kright) = (1-p_1)^k p_1, qquad k = 0,1,ldots$$
This also has an interpretation: this is you are counting the number of failures before success, so your definition is equivalent to
$$ widehat X = X-1.$$
From here we can determine the distribution of $Z = X + Y$ using the method you have
begin{align*}
mathbf{P}(Z = n) & = sum_{k=0}^n P(X = k) P(Y = n-k) \
& = sum_{k=1}^{n-1} P(X = k) P(Y = n-k) \
& = p_1 p_2 sum_{k=1}^{n-1} (1-p_1)^{k-1} (1-p_2)^{n-k-1} \
& = p_1 p_2frac{(1-p_2)^{n-1}}{(1-p_1)} sum_{k=1}^{n-1} left( frac{1-p_1}{1-p_2} right)^{k}.
end{align*}
From here you can manipulate the geometric series (much as you do above) to derive
$$mathbf{P}(Z = n) = frac{p_1p_2}{p_1 - p_2}
big( (1-p_2)^{n-1} - (1-p_1)^{n-1} big), qquad n = 2,3,ldots
$$
Note that $Z$ can only take values greater than or equal to $2$.
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
answered Feb 10 '18 at 9:38
owen88owen88
3,6721021
3,6721021
add a comment |
add a comment |
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2
$begingroup$
Tiny mistake - $a$ is $p_1 p_2 (1-p_2)^{n},$ not $p_1p_2$.
$endgroup$
– stochasticboy321
Feb 9 '18 at 22:14
$begingroup$
are you sure the question says density?
$endgroup$
– Henry
Feb 9 '18 at 22:54
$begingroup$
@Henry The book does say density but I believe the modern term probability mass function.
$endgroup$
– Bob
Feb 9 '18 at 23:03