Mixing
![What value does $frac{1}{n} sqrt[n]{n(n+1)(n+2)cdots(2n)}$ tend to? [duplicate]](https://lh5.googleusercontent.com/-9-r0a2TgAMc/AAAAAAAAAAI/AAAAAAAAG6s/4f97rFpP2A8/s72-c/photo.jpg?sz=32)
Find the distribution of $Z=X+Y$ where both $X$ and $Y$ are exponentially distributed.
$begingroup$
I have a problem where, in order to solve it, I need to find the distribution of $Z$. Say that $Xsimtext{exp}(lambda)$ and $Ysimtext{exp}(mu)$. I don't want to use the convolution formula but instead the mgf's. I have that
$$M_X(t)=frac{lambda}{lambda-t},quad quad M_Y(t)=frac{mu}{mu-t}.$$
Thus,
$$M_Z(t)=M_{X+Y}(t)=M_X(t)M_Y(t)=frac{lambdamu}{lambdamu-lambda t-mu t+t^2}.$$
I'm trying to find what distribution this is the mgf for. If $X$ and $Y$ were I.I.D, then it would just be an mgf for the gamma distribution. But this is not the case since I can't rewrite it in the correct form.
Is there any way to proceeed here?
probability probability-distributions generating-functions moment-generating-functions
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
$endgroup$
add a comment |
$begingroup$
I have a problem where, in order to solve it, I need to find the distribution of $Z$. Say that $Xsimtext{exp}(lambda)$ and $Ysimtext{exp}(mu)$. I don't want to use the convolution formula but instead the mgf's. I have that
$$M_X(t)=frac{lambda}{lambda-t},quad quad M_Y(t)=frac{mu}{mu-t}.$$
Thus,
$$M_Z(t)=M_{X+Y}(t)=M_X(t)M_Y(t)=frac{lambdamu}{lambdamu-lambda t-mu t+t^2}.$$
I'm trying to find what distribution this is the mgf for. If $X$ and $Y$ were I.I.D, then it would just be an mgf for the gamma distribution. But this is not the case since I can't rewrite it in the correct form.
Is there any way to proceeed here?
probability probability-distributions generating-functions moment-generating-functions
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
$endgroup$
$begingroup$
Are you assuming that $X$ and $Y$ are independent when you write $M_{X+Y}=M_XM_Y$?
$endgroup$
– Dante Grevino
Jan 9 at 23:10
$begingroup$
There isn't any particular name for this distribution, so far as I know. So the best you can do is present the MGF/CF, or compute the density directly via convolution.
$endgroup$
– Math1000
Jan 9 at 23:25
$begingroup$
@DanteGrevino Yes, independent but with different parameters.
$endgroup$
– Parseval
Jan 9 at 23:30
$begingroup$
@Math1000 - Thanks, good to know, then I don't need to try this anymore.
$endgroup$
– Parseval
Jan 9 at 23:31
add a comment |
$begingroup$
I have a problem where, in order to solve it, I need to find the distribution of $Z$. Say that $Xsimtext{exp}(lambda)$ and $Ysimtext{exp}(mu)$. I don't want to use the convolution formula but instead the mgf's. I have that
$$M_X(t)=frac{lambda}{lambda-t},quad quad M_Y(t)=frac{mu}{mu-t}.$$
Thus,
$$M_Z(t)=M_{X+Y}(t)=M_X(t)M_Y(t)=frac{lambdamu}{lambdamu-lambda t-mu t+t^2}.$$
I'm trying to find what distribution this is the mgf for. If $X$ and $Y$ were I.I.D, then it would just be an mgf for the gamma distribution. But this is not the case since I can't rewrite it in the correct form.
Is there any way to proceeed here?
probability probability-distributions generating-functions moment-generating-functions
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
$endgroup$
I have a problem where, in order to solve it, I need to find the distribution of $Z$. Say that $Xsimtext{exp}(lambda)$ and $Ysimtext{exp}(mu)$. I don't want to use the convolution formula but instead the mgf's. I have that
$$M_X(t)=frac{lambda}{lambda-t},quad quad M_Y(t)=frac{mu}{mu-t}.$$
Thus,
$$M_Z(t)=M_{X+Y}(t)=M_X(t)M_Y(t)=frac{lambdamu}{lambdamu-lambda t-mu t+t^2}.$$
I'm trying to find what distribution this is the mgf for. If $X$ and $Y$ were I.I.D, then it would just be an mgf for the gamma distribution. But this is not the case since I can't rewrite it in the correct form.
Is there any way to proceeed here?
probability probability-distributions generating-functions moment-generating-functions
probability probability-distributions generating-functions moment-generating-functions
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
asked Jan 9 at 22:30
ParsevalParseval
2,8261718
2,8261718
$begingroup$
Are you assuming that $X$ and $Y$ are independent when you write $M_{X+Y}=M_XM_Y$?
$endgroup$
– Dante Grevino
Jan 9 at 23:10
$begingroup$
There isn't any particular name for this distribution, so far as I know. So the best you can do is present the MGF/CF, or compute the density directly via convolution.
$endgroup$
– Math1000
Jan 9 at 23:25
$begingroup$
@DanteGrevino Yes, independent but with different parameters.
$endgroup$
– Parseval
Jan 9 at 23:30
$begingroup$
@Math1000 - Thanks, good to know, then I don't need to try this anymore.
$endgroup$
– Parseval
Jan 9 at 23:31
add a comment |
$begingroup$
Are you assuming that $X$ and $Y$ are independent when you write $M_{X+Y}=M_XM_Y$?
$endgroup$
– Dante Grevino
Jan 9 at 23:10
$begingroup$
There isn't any particular name for this distribution, so far as I know. So the best you can do is present the MGF/CF, or compute the density directly via convolution.
$endgroup$
– Math1000
Jan 9 at 23:25
$begingroup$
@DanteGrevino Yes, independent but with different parameters.
$endgroup$
– Parseval
Jan 9 at 23:30
$begingroup$
@Math1000 - Thanks, good to know, then I don't need to try this anymore.
$endgroup$
– Parseval
Jan 9 at 23:31
$begingroup$
Are you assuming that $X$ and $Y$ are independent when you write $M_{X+Y}=M_XM_Y$?
$endgroup$
– Dante Grevino
Jan 9 at 23:10
$begingroup$
Are you assuming that $X$ and $Y$ are independent when you write $M_{X+Y}=M_XM_Y$?
$endgroup$
– Dante Grevino
Jan 9 at 23:10
$begingroup$
There isn't any particular name for this distribution, so far as I know. So the best you can do is present the MGF/CF, or compute the density directly via convolution.
$endgroup$
– Math1000
Jan 9 at 23:25
$begingroup$
There isn't any particular name for this distribution, so far as I know. So the best you can do is present the MGF/CF, or compute the density directly via convolution.
$endgroup$
– Math1000
Jan 9 at 23:25
$begingroup$
@DanteGrevino Yes, independent but with different parameters.
$endgroup$
– Parseval
Jan 9 at 23:30
$begingroup$
@DanteGrevino Yes, independent but with different parameters.
$endgroup$
– Parseval
Jan 9 at 23:30
$begingroup$
@Math1000 - Thanks, good to know, then I don't need to try this anymore.
$endgroup$
– Parseval
Jan 9 at 23:31
$begingroup$
@Math1000 - Thanks, good to know, then I don't need to try this anymore.
$endgroup$
– Parseval
Jan 9 at 23:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{expo{-y/mu} over mu}
braces{bracks{z - y > 0}{expo{-pars{z - y}/lambda} over lambda}}dd y}
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
int_{0}^{z}exppars{{mu - lambda over lambdamu},y}dd y
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
{expo{pars{mu - lambda}z/pars{lambdamu}} -
1 over pars{mu - lambda}/pars{lambdamu}}dd y =
bbx{bracks{z > 0}{expo{-z/mu} - expo{-z/lambda} over mu - lambda}}
end{align}
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
$endgroup$
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
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{expo{-y/mu} over mu}
braces{bracks{z - y > 0}{expo{-pars{z - y}/lambda} over lambda}}dd y}
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
int_{0}^{z}exppars{{mu - lambda over lambdamu},y}dd y
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
{expo{pars{mu - lambda}z/pars{lambdamu}} -
1 over pars{mu - lambda}/pars{lambdamu}}dd y =
bbx{bracks{z > 0}{expo{-z/mu} - expo{-z/lambda} over mu - lambda}}
end{align}
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{expo{-y/mu} over mu}
braces{bracks{z - y > 0}{expo{-pars{z - y}/lambda} over lambda}}dd y}
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
int_{0}^{z}exppars{{mu - lambda over lambdamu},y}dd y
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
{expo{pars{mu - lambda}z/pars{lambdamu}} -
1 over pars{mu - lambda}/pars{lambdamu}}dd y =
bbx{bracks{z > 0}{expo{-z/mu} - expo{-z/lambda} over mu - lambda}}
end{align}
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
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newcommand{mc}[1]{mathcal{#1}}
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newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{expo{-y/mu} over mu}
braces{bracks{z - y > 0}{expo{-pars{z - y}/lambda} over lambda}}dd y}
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
int_{0}^{z}exppars{{mu - lambda over lambdamu},y}dd y
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
{expo{pars{mu - lambda}z/pars{lambdamu}} -
1 over pars{mu - lambda}/pars{lambdamu}}dd y =
bbx{bracks{z > 0}{expo{-z/mu} - expo{-z/lambda} over mu - lambda}}
end{align}
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{expo{-y/mu} over mu}
braces{bracks{z - y > 0}{expo{-pars{z - y}/lambda} over lambda}}dd y}
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
int_{0}^{z}exppars{{mu - lambda over lambdamu},y}dd y
\[5mm] = &
bracks{z > 0}{expo{-z/lambda} over lambdamu}
{expo{pars{mu - lambda}z/pars{lambdamu}} -
1 over pars{mu - lambda}/pars{lambdamu}}dd y =
bbx{bracks{z > 0}{expo{-z/mu} - expo{-z/lambda} over mu - lambda}}
end{align}
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
answered Jan 10 at 2:17
Felix MarinFelix Marin
67.8k7107142
67.8k7107142
add a comment |
add a comment |
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$begingroup$
Are you assuming that $X$ and $Y$ are independent when you write $M_{X+Y}=M_XM_Y$?
$endgroup$
– Dante Grevino
Jan 9 at 23:10
$begingroup$
There isn't any particular name for this distribution, so far as I know. So the best you can do is present the MGF/CF, or compute the density directly via convolution.
$endgroup$
– Math1000
Jan 9 at 23:25
$begingroup$
@DanteGrevino Yes, independent but with different parameters.
$endgroup$
– Parseval
Jan 9 at 23:30
$begingroup$
@Math1000 - Thanks, good to know, then I don't need to try this anymore.
$endgroup$
– Parseval
Jan 9 at 23:31