Existence of a limiting sum of random variables
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Consider that $X_i$‘s are independent exponentially distributed random variables with mean $1/i$ (and thus variance $1/i^2$).Then the sum of them seems to converge to a “random variable” with finite variance but unbounded mean. What’s the problem here? Why does not such random variable exist?
Thanks
probability statistics random-variables
asked Nov 14 at 22:35
user49229
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up vote
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Consider that $X_i$‘s are independent exponentially distributed random variables with mean $1/i$ (and thus variance $1/i^2$).Then the sum of them seems to converge to a “random variable” with finite variance but unbounded mean. What’s the problem here? Why does not such random variable exist?
Thanks
probability statistics random-variables
asked Nov 14 at 22:35
user49229
144
1
By the same reasoning the "random variable" sequence ${Y_i}_{i=1}^{infty}$ defined by $Y_i=i$ for all $i$ (with prob 1) has partial sums $sum_{i=1}^n Y_i$ with zero variance but mean that goes to infinity. But there is no mystery in this. We certainly do not say that $sum_{i=1}^n i$ "converges" in any sense (it diverges to $infty$). [I put "random variable" in quotes simply because the ${Y_i}$ sequence is deterministic in this case.]
– Michael
Nov 14 at 22:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider that $X_i$‘s are independent exponentially distributed random variables with mean $1/i$ (and thus variance $1/i^2$).Then the sum of them seems to converge to a “random variable” with finite variance but unbounded mean. What’s the problem here? Why does not such random variable exist?
Thanks
probability statistics random-variables
asked Nov 14 at 22:35
user49229
144
Consider that $X_i$‘s are independent exponentially distributed random variables with mean $1/i$ (and thus variance $1/i^2$).Then the sum of them seems to converge to a “random variable” with finite variance but unbounded mean. What’s the problem here? Why does not such random variable exist?
Thanks
probability statistics random-variables
probability statistics random-variables
asked Nov 14 at 22:35
user49229
144
asked Nov 14 at 22:35
user49229
144
edited Nov 15 at 3:04
asked Nov 14 at 22:35
user49229
144
asked Nov 14 at 22:35
user49229
144
asked Nov 14 at 22:35
user49229
144
144
1
By the same reasoning the "random variable" sequence ${Y_i}_{i=1}^{infty}$ defined by $Y_i=i$ for all $i$ (with prob 1) has partial sums $sum_{i=1}^n Y_i$ with zero variance but mean that goes to infinity. But there is no mystery in this. We certainly do not say that $sum_{i=1}^n i$ "converges" in any sense (it diverges to $infty$). [I put "random variable" in quotes simply because the ${Y_i}$ sequence is deterministic in this case.]
– Michael
Nov 14 at 22:49
add a comment |
1
By the same reasoning the "random variable" sequence ${Y_i}_{i=1}^{infty}$ defined by $Y_i=i$ for all $i$ (with prob 1) has partial sums $sum_{i=1}^n Y_i$ with zero variance but mean that goes to infinity. But there is no mystery in this. We certainly do not say that $sum_{i=1}^n i$ "converges" in any sense (it diverges to $infty$). [I put "random variable" in quotes simply because the ${Y_i}$ sequence is deterministic in this case.]
– Michael
Nov 14 at 22:49
1
1
By the same reasoning the "random variable" sequence ${Y_i}_{i=1}^{infty}$ defined by $Y_i=i$ for all $i$ (with prob 1) has partial sums $sum_{i=1}^n Y_i$ with zero variance but mean that goes to infinity. But there is no mystery in this. We certainly do not say that $sum_{i=1}^n i$ "converges" in any sense (it diverges to $infty$). [I put "random variable" in quotes simply because the ${Y_i}$ sequence is deterministic in this case.]
– Michael
Nov 14 at 22:49
By the same reasoning the "random variable" sequence ${Y_i}_{i=1}^{infty}$ defined by $Y_i=i$ for all $i$ (with prob 1) has partial sums $sum_{i=1}^n Y_i$ with zero variance but mean that goes to infinity. But there is no mystery in this. We certainly do not say that $sum_{i=1}^n i$ "converges" in any sense (it diverges to $infty$). [I put "random variable" in quotes simply because the ${Y_i}$ sequence is deterministic in this case.]
– Michael
Nov 14 at 22:49
add a comment |
1 Answer
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Since $X_i$'s are positive $sum_{i=1}^{n} X_i$ converges (to a possibly infinte ) sum $X$ and Monotone Convergence Theorem gives $EX=sum_{i=1}^{infty} frac 1 i =infty$. There is no way a random variable with infinite mean can have finite variance. It is not even true that $X <infty$ almost surely. [This can be shown using Laplace transforms].
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since $X_i$'s are positive $sum_{i=1}^{n} X_i$ converges (to a possibly infinte ) sum $X$ and Monotone Convergence Theorem gives $EX=sum_{i=1}^{infty} frac 1 i =infty$. There is no way a random variable with infinite mean can have finite variance. It is not even true that $X <infty$ almost surely. [This can be shown using Laplace transforms].
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
add a comment |
up vote
1
down vote
Since $X_i$'s are positive $sum_{i=1}^{n} X_i$ converges (to a possibly infinte ) sum $X$ and Monotone Convergence Theorem gives $EX=sum_{i=1}^{infty} frac 1 i =infty$. There is no way a random variable with infinite mean can have finite variance. It is not even true that $X <infty$ almost surely. [This can be shown using Laplace transforms].
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $X_i$'s are positive $sum_{i=1}^{n} X_i$ converges (to a possibly infinte ) sum $X$ and Monotone Convergence Theorem gives $EX=sum_{i=1}^{infty} frac 1 i =infty$. There is no way a random variable with infinite mean can have finite variance. It is not even true that $X <infty$ almost surely. [This can be shown using Laplace transforms].
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
Since $X_i$'s are positive $sum_{i=1}^{n} X_i$ converges (to a possibly infinte ) sum $X$ and Monotone Convergence Theorem gives $EX=sum_{i=1}^{infty} frac 1 i =infty$. There is no way a random variable with infinite mean can have finite variance. It is not even true that $X <infty$ almost surely. [This can be shown using Laplace transforms].
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
edited Nov 14 at 23:35
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
answered Nov 14 at 23:30
Kavi Rama Murthy
41.3k31751
41.3k31751
add a comment |
add a comment |
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1
By the same reasoning the "random variable" sequence ${Y_i}_{i=1}^{infty}$ defined by $Y_i=i$ for all $i$ (with prob 1) has partial sums $sum_{i=1}^n Y_i$ with zero variance but mean that goes to infinity. But there is no mystery in this. We certainly do not say that $sum_{i=1}^n i$ "converges" in any sense (it diverges to $infty$). [I put "random variable" in quotes simply because the ${Y_i}$ sequence is deterministic in this case.]
– Michael
Nov 14 at 22:49