Distribution of the sum of n independent variables of the exponential family.
up vote
1
down vote
favorite
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 at 23:31
PedroGonçalves
386
add a comment |
up vote
1
down vote
favorite
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 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 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 at 14:43
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 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 at 23:31
PedroGonçalves
386
asked Nov 14 at 23:31
PedroGonçalves
386
asked Nov 14 at 23:31
PedroGonçalves
386
asked Nov 14 at 23:31
PedroGonçalves
386
asked Nov 14 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 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 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 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 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 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 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 at 14:42
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 at 14:43
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 at 14:43
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
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 2 days ago
Aditya Dua
4906
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
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 2 days ago
Aditya Dua
4906
add a comment |
up vote
0
down vote
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 2 days ago
Aditya Dua
4906
add a comment |
up vote
0
down vote
up vote
0
down vote
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 2 days ago
Aditya Dua
4906
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 2 days ago
Aditya Dua
4906
answered 2 days ago
Aditya Dua
4906
answered 2 days ago
Aditya Dua
4906
answered 2 days ago
Aditya Dua
4906
4906
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998967%2fdistribution-of-the-sum-of-n-independent-variables-of-the-exponential-family%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 at 0:19
Hard to tell without being given the exact distribution of the $Y_i$'s.
– StubbornAtom
Nov 16 at 14:42
@BruceET Question says "exponential family".
– StubbornAtom
Nov 16 at 14:43