Mixing
Distribution of the sum of n independent variables of the exponential family.
Suppose you have $n$ random and independent variables $Y_{1},...,Y_{n}$ whose distribution belongs to the uniparametric exponential family.
How do I find the distribution of $sum_{i=1}^{n} Yi$ ?
statistics probability-distributions statistical-inference
asked Nov 14 '18 at 23:31
PedroGonçalves
386
add a comment |
Suppose you have $n$ random and independent variables $Y_{1},...,Y_{n}$ whose distribution belongs to the uniparametric exponential family.
How do I find the distribution of $sum_{i=1}^{n} Yi$ ?
statistics probability-distributions statistical-inference
asked Nov 14 '18 at 23:31
PedroGonçalves
386
Suggest you use moment generating functions; the sum of $n$ iid exponential random variables is distributed as a gamma distribution with shape parameter $n.$ // Some Answers listed at the right under 'Related' may be helpful.
– BruceET
Nov 15 '18 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 '18 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 '18 at 14:43
add a comment |
Suppose you have $n$ random and independent variables $Y_{1},...,Y_{n}$ whose distribution belongs to the uniparametric exponential family.
How do I find the distribution of $sum_{i=1}^{n} Yi$ ?
statistics probability-distributions statistical-inference
asked Nov 14 '18 at 23:31
PedroGonçalves
386
Suppose you have $n$ random and independent variables $Y_{1},...,Y_{n}$ whose distribution belongs to the uniparametric exponential family.
How do I find the distribution of $sum_{i=1}^{n} Yi$ ?
statistics probability-distributions statistical-inference
statistics probability-distributions statistical-inference
asked Nov 14 '18 at 23:31
PedroGonçalves
386
asked Nov 14 '18 at 23:31
PedroGonçalves
386
asked Nov 14 '18 at 23:31
PedroGonçalves
386
asked Nov 14 '18 at 23:31
PedroGonçalves
386
asked Nov 14 '18 at 23:31
PedroGonçalves
386
386
Suggest you use moment generating functions; the sum of $n$ iid exponential random variables is distributed as a gamma distribution with shape parameter $n.$ // Some Answers listed at the right under 'Related' may be helpful.
– BruceET
Nov 15 '18 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 '18 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 '18 at 14:43
add a comment |
Suggest you use moment generating functions; the sum of $n$ iid exponential random variables is distributed as a gamma distribution with shape parameter $n.$ // Some Answers listed at the right under 'Related' may be helpful.
– BruceET
Nov 15 '18 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 '18 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 '18 at 14:43
Suggest you use moment generating functions; the sum of $n$ iid exponential random variables is distributed as a gamma distribution with shape parameter $n.$ // Some Answers listed at the right under 'Related' may be helpful.
– BruceET
Nov 15 '18 at 0:19
Suggest you use moment generating functions; the sum of $n$ iid exponential random variables is distributed as a gamma distribution with shape parameter $n.$ // Some Answers listed at the right under 'Related' may be helpful.
– BruceET
Nov 15 '18 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 '18 at 14:42
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 '18 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 '18 at 14:43
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 '18 at 14:43
add a comment |
1 Answer
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Let $Y_i sim exp(1)$, i.e. $f_Y(y) = e^{-y}$ and $F_y(y)=1-e^{-y}$. You can extend the analysis easily to $Y_i sim exp(lambda)$.
Let's consider the case of $n=2$, i.e. $X = Y_1+Y_2$.
$F_X(x) = P(X leq x)$
= $P(Y_1+Y_2 leq x)$
= $int_0^x P(Y_1 leq x-y) f_Y(y) dy$
= $int_0^x left( 1-e^{-(x-y)}right) e^{-y} dy$
= $int_0^x (e^{-y} - e^{-x})dy$
= $1-e^{-x}-xe^{-x}$
You can differentiate to get the PDF $f_X(x) = xe^{-x}$.
If you do this a couple more times, you will see a pattern, at which point you can arrive at the answer by the principle of mathematical induction. The moment generating function approach is of course much quicker, if you are familiar with that.
answered Nov 21 '18 at 6:08
Aditya Dua
86418
add a comment |
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1 Answer
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1 Answer
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Let $Y_i sim exp(1)$, i.e. $f_Y(y) = e^{-y}$ and $F_y(y)=1-e^{-y}$. You can extend the analysis easily to $Y_i sim exp(lambda)$.
Let's consider the case of $n=2$, i.e. $X = Y_1+Y_2$.
$F_X(x) = P(X leq x)$
= $P(Y_1+Y_2 leq x)$
= $int_0^x P(Y_1 leq x-y) f_Y(y) dy$
= $int_0^x left( 1-e^{-(x-y)}right) e^{-y} dy$
= $int_0^x (e^{-y} - e^{-x})dy$
= $1-e^{-x}-xe^{-x}$
You can differentiate to get the PDF $f_X(x) = xe^{-x}$.
If you do this a couple more times, you will see a pattern, at which point you can arrive at the answer by the principle of mathematical induction. The moment generating function approach is of course much quicker, if you are familiar with that.
answered Nov 21 '18 at 6:08
Aditya Dua
86418
add a comment |
Let $Y_i sim exp(1)$, i.e. $f_Y(y) = e^{-y}$ and $F_y(y)=1-e^{-y}$. You can extend the analysis easily to $Y_i sim exp(lambda)$.
Let's consider the case of $n=2$, i.e. $X = Y_1+Y_2$.
$F_X(x) = P(X leq x)$
= $P(Y_1+Y_2 leq x)$
= $int_0^x P(Y_1 leq x-y) f_Y(y) dy$
= $int_0^x left( 1-e^{-(x-y)}right) e^{-y} dy$
= $int_0^x (e^{-y} - e^{-x})dy$
= $1-e^{-x}-xe^{-x}$
You can differentiate to get the PDF $f_X(x) = xe^{-x}$.
If you do this a couple more times, you will see a pattern, at which point you can arrive at the answer by the principle of mathematical induction. The moment generating function approach is of course much quicker, if you are familiar with that.
answered Nov 21 '18 at 6:08
Aditya Dua
86418
add a comment |
Let $Y_i sim exp(1)$, i.e. $f_Y(y) = e^{-y}$ and $F_y(y)=1-e^{-y}$. You can extend the analysis easily to $Y_i sim exp(lambda)$.
Let's consider the case of $n=2$, i.e. $X = Y_1+Y_2$.
$F_X(x) = P(X leq x)$
= $P(Y_1+Y_2 leq x)$
= $int_0^x P(Y_1 leq x-y) f_Y(y) dy$
= $int_0^x left( 1-e^{-(x-y)}right) e^{-y} dy$
= $int_0^x (e^{-y} - e^{-x})dy$
= $1-e^{-x}-xe^{-x}$
You can differentiate to get the PDF $f_X(x) = xe^{-x}$.
If you do this a couple more times, you will see a pattern, at which point you can arrive at the answer by the principle of mathematical induction. The moment generating function approach is of course much quicker, if you are familiar with that.
answered Nov 21 '18 at 6:08
Aditya Dua
86418
Let $Y_i sim exp(1)$, i.e. $f_Y(y) = e^{-y}$ and $F_y(y)=1-e^{-y}$. You can extend the analysis easily to $Y_i sim exp(lambda)$.
Let's consider the case of $n=2$, i.e. $X = Y_1+Y_2$.
$F_X(x) = P(X leq x)$
= $P(Y_1+Y_2 leq x)$
= $int_0^x P(Y_1 leq x-y) f_Y(y) dy$
= $int_0^x left( 1-e^{-(x-y)}right) e^{-y} dy$
= $int_0^x (e^{-y} - e^{-x})dy$
= $1-e^{-x}-xe^{-x}$
You can differentiate to get the PDF $f_X(x) = xe^{-x}$.
If you do this a couple more times, you will see a pattern, at which point you can arrive at the answer by the principle of mathematical induction. The moment generating function approach is of course much quicker, if you are familiar with that.
answered Nov 21 '18 at 6:08
Aditya Dua
86418
answered Nov 21 '18 at 6:08
Aditya Dua
86418
answered Nov 21 '18 at 6:08
Aditya Dua
86418
answered Nov 21 '18 at 6:08
Aditya Dua
86418
86418
add a comment |
add a comment |
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Suggest you use moment generating functions; the sum of $n$ iid exponential random variables is distributed as a gamma distribution with shape parameter $n.$ // Some Answers listed at the right under 'Related' may be helpful.
– BruceET
Nov 15 '18 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 '18 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 '18 at 14:43