Mixing
Convergence of sequence of uniforms
$begingroup$
Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.
I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$
Then:
$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$
As n goes to infinity, I think I get the following function:
$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?
To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.
probability convergence random-variables weak-convergence uniform-distribution
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
asked Jan 18 at 16:11
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.
I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$
Then:
$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$
As n goes to infinity, I think I get the following function:
$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?
To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.
probability convergence random-variables weak-convergence uniform-distribution
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
asked Jan 18 at 16:11
qcc101qcc101
627213
$endgroup$
$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24
add a comment |
$begingroup$
Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.
I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$
Then:
$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$
As n goes to infinity, I think I get the following function:
$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?
To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.
probability convergence random-variables weak-convergence uniform-distribution
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
asked Jan 18 at 16:11
qcc101qcc101
627213
$endgroup$
Let $Xsimmathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.
I did it using the distribution function, I have:
$$F_X(x) = x mathbb{1}_{[0,1](x)} + mathbb{1}_{[1,infty](x)}$$
Then:
$$F_{X_n}(x) = x^{frac{1}{n}}mathbb{1}_{[0,1]}(x) + mathbb{1}_{[1,infty]}(x)$$
As n goes to infinity, I think I get the following function:
$$G(x) = mathbb{1}_{[0,infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n to Y$ in law. This may sound stupid, but which random variable has such distribution?
To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.
probability convergence random-variables weak-convergence uniform-distribution
probability convergence random-variables weak-convergence uniform-distribution
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
asked Jan 18 at 16:11
qcc101qcc101
627213
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
asked Jan 18 at 16:11
qcc101qcc101
627213
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
edited Jan 20 at 13:40
Davide Giraudo
127k16151265
127k16151265
asked Jan 18 at 16:11
qcc101qcc101
627213
asked Jan 18 at 16:11
qcc101qcc101
627213
asked Jan 18 at 16:11
qcc101qcc101
627213
627213
$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24
add a comment |
$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24
$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24
$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.
Actually, the convergence $X_nto 0$ holds almost surely hence in probability
hence in distribution.
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
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$begingroup$
Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.
Actually, the convergence $X_nto 0$ holds almost surely hence in probability
hence in distribution.
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
$begingroup$
Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.
Actually, the convergence $X_nto 0$ holds almost surely hence in probability
hence in distribution.
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
add a comment |
$begingroup$
Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.
Actually, the convergence $X_nto 0$ holds almost surely hence in probability
hence in distribution.
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
$endgroup$
Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.
Actually, the convergence $X_nto 0$ holds almost surely hence in probability
hence in distribution.
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
edited Jan 24 at 10:21
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
answered Jan 19 at 10:53
Davide GiraudoDavide Giraudo
127k16151265
127k16151265
add a comment |
add a comment |
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$begingroup$
Yes. $F(Y) = mathbb{1}_{[0,infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $mathbb{1}_{(0,infty)(y)}$ but that is not a CDF as it is not càdlàg
$endgroup$
– Henry
Jan 18 at 16:24