Mixing
Limiting behavior of sum of uniformly distributed random variables
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I was looking into the following problem and I can not think of a solution.
Consider i.i.d. random variables $X_1, X_2, ...$, such that $X_i$ has a uniform distribution over the interval $(0, 1)$. Find the following limit:
$$lim_n to infty Pleft(sum_i=1^n X_i leq 2 sum_i=1^nX_i^2right).$$
The only thing that occurred to me is evaluate each summand independently when
$$X_i leq 2 X^2_i$$ is equivalent to $X_i(1-X_i) leq X^2_i$, which implies $(1-X_i leq X_i)$ and the probability of this event is equal to $1/2$. Overall probability would be $(1/2)^n$ (since the summands are independent), its limit would be $0$. But that is not the limit I was supposed to look for.
probability-theory uniform-distribution
edited Aug 2 at 15:51
Math1000
18.4k31444
asked Aug 2 at 15:46
Vwann
224
add a comment |Â
up vote
2
down vote
favorite
I was looking into the following problem and I can not think of a solution.
Consider i.i.d. random variables $X_1, X_2, ...$, such that $X_i$ has a uniform distribution over the interval $(0, 1)$. Find the following limit:
$$lim_n to infty Pleft(sum_i=1^n X_i leq 2 sum_i=1^nX_i^2right).$$
The only thing that occurred to me is evaluate each summand independently when
$$X_i leq 2 X^2_i$$ is equivalent to $X_i(1-X_i) leq X^2_i$, which implies $(1-X_i leq X_i)$ and the probability of this event is equal to $1/2$. Overall probability would be $(1/2)^n$ (since the summands are independent), its limit would be $0$. But that is not the limit I was supposed to look for.
probability-theory uniform-distribution
edited Aug 2 at 15:51
Math1000
18.4k31444
asked Aug 2 at 15:46
Vwann
224
Yes, but it's not required that all the $X_igeq frac12$ in order for the inequality to hold.
– saulspatz
Aug 2 at 15:50
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was looking into the following problem and I can not think of a solution.
Consider i.i.d. random variables $X_1, X_2, ...$, such that $X_i$ has a uniform distribution over the interval $(0, 1)$. Find the following limit:
$$lim_n to infty Pleft(sum_i=1^n X_i leq 2 sum_i=1^nX_i^2right).$$
The only thing that occurred to me is evaluate each summand independently when
$$X_i leq 2 X^2_i$$ is equivalent to $X_i(1-X_i) leq X^2_i$, which implies $(1-X_i leq X_i)$ and the probability of this event is equal to $1/2$. Overall probability would be $(1/2)^n$ (since the summands are independent), its limit would be $0$. But that is not the limit I was supposed to look for.
probability-theory uniform-distribution
edited Aug 2 at 15:51
Math1000
18.4k31444
asked Aug 2 at 15:46
Vwann
224
I was looking into the following problem and I can not think of a solution.
Consider i.i.d. random variables $X_1, X_2, ...$, such that $X_i$ has a uniform distribution over the interval $(0, 1)$. Find the following limit:
$$lim_n to infty Pleft(sum_i=1^n X_i leq 2 sum_i=1^nX_i^2right).$$
The only thing that occurred to me is evaluate each summand independently when
$$X_i leq 2 X^2_i$$ is equivalent to $X_i(1-X_i) leq X^2_i$, which implies $(1-X_i leq X_i)$ and the probability of this event is equal to $1/2$. Overall probability would be $(1/2)^n$ (since the summands are independent), its limit would be $0$. But that is not the limit I was supposed to look for.
probability-theory uniform-distribution
edited Aug 2 at 15:51
Math1000
18.4k31444
asked Aug 2 at 15:46
Vwann
224
edited Aug 2 at 15:51
Math1000
18.4k31444
edited Aug 2 at 15:51
Math1000
18.4k31444
edited Aug 2 at 15:51
Math1000
18.4k31444
18.4k31444
asked Aug 2 at 15:46
Vwann
224
asked Aug 2 at 15:46
Vwann
224
asked Aug 2 at 15:46
Vwann
224
224
Yes, but it's not required that all the $X_igeq frac12$ in order for the inequality to hold.
– saulspatz
Aug 2 at 15:50
add a comment |Â
Yes, but it's not required that all the $X_igeq frac12$ in order for the inequality to hold.
– saulspatz
Aug 2 at 15:50
Yes, but it's not required that all the $X_igeq frac12$ in order for the inequality to hold.
– saulspatz
Aug 2 at 15:50
Yes, but it's not required that all the $X_igeq frac12$ in order for the inequality to hold.
– saulspatz
Aug 2 at 15:50
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint: divide both sides by $n$. What does the strong law of large numbers say about each side as $n to infty$?
answered Aug 2 at 15:58
Marcus M
8,1631847
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: divide both sides by $n$. What does the strong law of large numbers say about each side as $n to infty$?
answered Aug 2 at 15:58
Marcus M
8,1631847
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
add a comment |Â
up vote
1
down vote
accepted
Hint: divide both sides by $n$. What does the strong law of large numbers say about each side as $n to infty$?
answered Aug 2 at 15:58
Marcus M
8,1631847
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: divide both sides by $n$. What does the strong law of large numbers say about each side as $n to infty$?
answered Aug 2 at 15:58
Marcus M
8,1631847
Hint: divide both sides by $n$. What does the strong law of large numbers say about each side as $n to infty$?
answered Aug 2 at 15:58
Marcus M
8,1631847
answered Aug 2 at 15:58
Marcus M
8,1631847
answered Aug 2 at 15:58
Marcus M
8,1631847
answered Aug 2 at 15:58
Marcus M
8,1631847
8,1631847
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
add a comment |Â
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
That the left side converges to the mean of a uniformly distributed r.v. which is $1/2$ and the right side converges to twice the mean of a squared unif. distributed r.v. which is $2/3$ in this case.
– Vwann
Aug 2 at 16:06
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
Wouldn't the weak law of large numbers be more applicable here?
– Math1000
Aug 2 at 16:11
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
So can I state that since $$P(lim_n to infty hatX^_n - 2hatX^2_n = -1/6) =1$$ (convergence almost surely) then the original limit is 1 also?
– Vwann
Aug 2 at 16:13
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
Yes, the original limit is $1$ also; you could use the weak law of large numbers too, but that would require a tiny bit more work. It's basically the same thing.
– Marcus M
Aug 2 at 16:24
add a comment |Â
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Yes, but it's not required that all the $X_igeq frac12$ in order for the inequality to hold.
– saulspatz
Aug 2 at 15:50