Sequence of random variables and convergence in probability.
$begingroup$
Let $Omega:=[0,1]$ with Lebesgue measure.Let's define a sequence $X_n= frac{omega}{n}$.
Check:
a) Convergence in distribution
b) Convergence in probability
My solution:
b)
$P(|X_n-X| ge epsilon)=P(|X_n| ge epsilon)=P(frac{omega}{n} ge epsilon)=P(omega ge epsilon n )$
And this is equal 0 for each $epsilon >0$ and $ n to infty$.
a)
$X equiv 0$
We have $F_X(t)= begin{cases} 0 &text{for } t < 0\1 &text{for } t ge 0end{cases}$
Now we have to calculate $F_{X_n}$ and here I have problem.
probability convergence random-variables
asked Jan 17 at 16:36
pawelKpawelK
618
$endgroup$
add a comment |
$begingroup$
Let $Omega:=[0,1]$ with Lebesgue measure.Let's define a sequence $X_n= frac{omega}{n}$.
Check:
a) Convergence in distribution
b) Convergence in probability
My solution:
b)
$P(|X_n-X| ge epsilon)=P(|X_n| ge epsilon)=P(frac{omega}{n} ge epsilon)=P(omega ge epsilon n )$
And this is equal 0 for each $epsilon >0$ and $ n to infty$.
a)
$X equiv 0$
We have $F_X(t)= begin{cases} 0 &text{for } t < 0\1 &text{for } t ge 0end{cases}$
Now we have to calculate $F_{X_n}$ and here I have problem.
probability convergence random-variables
asked Jan 17 at 16:36
pawelKpawelK
618
$endgroup$
add a comment |
$begingroup$
Let $Omega:=[0,1]$ with Lebesgue measure.Let's define a sequence $X_n= frac{omega}{n}$.
Check:
a) Convergence in distribution
b) Convergence in probability
My solution:
b)
$P(|X_n-X| ge epsilon)=P(|X_n| ge epsilon)=P(frac{omega}{n} ge epsilon)=P(omega ge epsilon n )$
And this is equal 0 for each $epsilon >0$ and $ n to infty$.
a)
$X equiv 0$
We have $F_X(t)= begin{cases} 0 &text{for } t < 0\1 &text{for } t ge 0end{cases}$
Now we have to calculate $F_{X_n}$ and here I have problem.
probability convergence random-variables
asked Jan 17 at 16:36
pawelKpawelK
618
$endgroup$
Let $Omega:=[0,1]$ with Lebesgue measure.Let's define a sequence $X_n= frac{omega}{n}$.
Check:
a) Convergence in distribution
b) Convergence in probability
My solution:
b)
$P(|X_n-X| ge epsilon)=P(|X_n| ge epsilon)=P(frac{omega}{n} ge epsilon)=P(omega ge epsilon n )$
And this is equal 0 for each $epsilon >0$ and $ n to infty$.
a)
$X equiv 0$
We have $F_X(t)= begin{cases} 0 &text{for } t < 0\1 &text{for } t ge 0end{cases}$
Now we have to calculate $F_{X_n}$ and here I have problem.
probability convergence random-variables
probability convergence random-variables
asked Jan 17 at 16:36
pawelKpawelK
618
asked Jan 17 at 16:36
pawelKpawelK
618
asked Jan 17 at 16:36
pawelKpawelK
618
asked Jan 17 at 16:36
pawelKpawelK
618
asked Jan 17 at 16:36
pawelKpawelK
618
618
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Convergence in probability implies convergence in distribution, so having done b) you do not need to do a).
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Convergence in probability implies convergence in distribution, so having done b) you do not need to do a).
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
add a comment |
$begingroup$
Convergence in probability implies convergence in distribution, so having done b) you do not need to do a).
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
add a comment |
$begingroup$
Convergence in probability implies convergence in distribution, so having done b) you do not need to do a).
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
$endgroup$
Convergence in probability implies convergence in distribution, so having done b) you do not need to do a).
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
answered Jan 17 at 18:59
Jakub AndruszkiewiczJakub Andruszkiewicz
2116
2116
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
add a comment |
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
Ok, but I have problem with a.
$endgroup$
– pawelK
Jan 17 at 19:06
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
$begingroup$
You shouldn't have, as you did b) and b) => a). But ok, if you want to do it without using b), then you can see that F_X_n(t) = 0 for t<0, 1 for t>1/n, and it is linear on [0,1/n], so nt for t in [0,1/n].
$endgroup$
– Jakub Andruszkiewicz
Jan 17 at 19:12
add a comment |
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