Convergence in probability
$begingroup$
Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.
Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$
probability sequences-and-series convergence random-variables uniform-distribution
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
$endgroup$
add a comment |
$begingroup$
Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.
Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$
probability sequences-and-series convergence random-variables uniform-distribution
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
$endgroup$
add a comment |
$begingroup$
Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.
Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$
probability sequences-and-series convergence random-variables uniform-distribution
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
$endgroup$
Can anyone tell me how they got the regions $0<epsilon<theta$ and $epsilon >0 $.
Also to clarify, is the last step where it says $lim_{n to infty} P(|Y_n-theta|>epsilon)=0$
probability sequences-and-series convergence random-variables uniform-distribution
probability sequences-and-series convergence random-variables uniform-distribution
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
edited Dec 17 '18 at 22:54
Davide Giraudo
127k17154268
127k17154268
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
asked Jun 5 '16 at 14:23
HueHueHueHue
12310
12310
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
begin{align}
mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
&= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
&= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
&= F_{Y_n}(theta- epsilon).
end{align}
Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:
(1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
$$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
as $n to infty$.
(2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.
answered Jun 5 '16 at 15:42
SironSiron
1,205614
$endgroup$
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
1
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
add a comment |
$begingroup$
Imagine we have $X$ representing the toss of a fair die.
Then $text{Range}(X) = {1,2,cdots,6}$
$$P(X = x) = frac16$$
So:
$$P(X le 1) = frac16$$
$$P(X le 2) = frac26$$
$$P(X le 3) = frac36$$
$$P(X le 4) = frac46$$
$$P(X le 5) = frac56$$
$$P(X le 6) = frac66 = 1$$
Also:
$$P(X le 7) = frac66 = 1$$
$$P(X le 0) = frac06$$
$$P(X le 0.5) = frac06$$
$$P(X le -2000) = frac06$$
$$P(X le 4.5) = frac46$$
$$P(X > 7) = frac06$$
Thus:
The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'
The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'
Also
Show that $Y_n$ converges in probability
edited Apr 13 '17 at 12:19
Community♦
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
begin{align}
mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
&= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
&= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
&= F_{Y_n}(theta- epsilon).
end{align}
Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:
(1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
$$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
as $n to infty$.
(2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.
answered Jun 5 '16 at 15:42
SironSiron
1,205614
$endgroup$
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
1
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
add a comment |
$begingroup$
begin{align}
mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
&= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
&= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
&= F_{Y_n}(theta- epsilon).
end{align}
Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:
(1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
$$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
as $n to infty$.
(2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.
answered Jun 5 '16 at 15:42
SironSiron
1,205614
$endgroup$
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
1
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
add a comment |
$begingroup$
begin{align}
mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
&= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
&= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
&= F_{Y_n}(theta- epsilon).
end{align}
Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:
(1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
$$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
as $n to infty$.
(2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.
answered Jun 5 '16 at 15:42
SironSiron
1,205614
$endgroup$
begin{align}
mathbb{P}(|Y_n - theta | geq epsilon) &= 1- mathbb{P}(-epsilon leq Y_n - theta leq epsilon) \
&= 1 - mathbb{P}(Y_n leq theta + epsilon) + mathbb{P}(Y_n leq theta - epsilon) \
&= 1 - F_{Y_n}(theta + epsilon) + F_{Y_n}(theta-epsilon). \
&= F_{Y_n}(theta- epsilon).
end{align}
Note that the last equality follows from the fact that $theta + epsilon geq theta$ for any $epsilon>0$ and consequently $F_{Y_n}(theta+ epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:
(1) Assume $0<epsilon<theta$, then $0 < theta-epsilon < theta$ and consequently
$$mathbb{P}(|Y_n - theta| geq epsilon) = left(frac{theta-epsilon}{theta}right)^n to 0,$$
as $n to infty$.
(2) Assume $epsilon > theta$, then $theta - epsilon < 0$ and consequently $F_{Y_n}(theta- epsilon) = 0, forall n geq 1$. Therefore, the result holds in the limit.
answered Jun 5 '16 at 15:42
SironSiron
1,205614
edited Dec 17 '18 at 22:18
answered Jun 5 '16 at 15:42
SironSiron
1,205614
answered Jun 5 '16 at 15:42
SironSiron
1,205614
answered Jun 5 '16 at 15:42
SironSiron
1,205614
1,205614
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
1
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
add a comment |
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
1
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
$begingroup$
I'm a little confused with your statement, Shouldn't $F_{yn}(theta -epsilon)=(frac{theta -epsilon}{theta})^{n}$
$endgroup$
– HueHue
Jun 5 '16 at 16:14
1
1
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Yes ... and $mathbb{P}(|Y_n -theta | geq epsilon) = F_{Y_n}(theta-epsilon)$ as showed above.
$endgroup$
– Siron
Jun 5 '16 at 16:26
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
Sorry I meant I was confused with the bit where you said "consider $epsilon geq theta$. In that case..."
$endgroup$
– HueHue
Jun 5 '16 at 16:31
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
$begingroup$
What exactly are you confused about?
$endgroup$
– Siron
Jun 5 '16 at 16:45
add a comment |
$begingroup$
Imagine we have $X$ representing the toss of a fair die.
Then $text{Range}(X) = {1,2,cdots,6}$
$$P(X = x) = frac16$$
So:
$$P(X le 1) = frac16$$
$$P(X le 2) = frac26$$
$$P(X le 3) = frac36$$
$$P(X le 4) = frac46$$
$$P(X le 5) = frac56$$
$$P(X le 6) = frac66 = 1$$
Also:
$$P(X le 7) = frac66 = 1$$
$$P(X le 0) = frac06$$
$$P(X le 0.5) = frac06$$
$$P(X le -2000) = frac06$$
$$P(X le 4.5) = frac46$$
$$P(X > 7) = frac06$$
Thus:
The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'
The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'
Also
Show that $Y_n$ converges in probability
edited Apr 13 '17 at 12:19
Community♦
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
$endgroup$
add a comment |
$begingroup$
Imagine we have $X$ representing the toss of a fair die.
Then $text{Range}(X) = {1,2,cdots,6}$
$$P(X = x) = frac16$$
So:
$$P(X le 1) = frac16$$
$$P(X le 2) = frac26$$
$$P(X le 3) = frac36$$
$$P(X le 4) = frac46$$
$$P(X le 5) = frac56$$
$$P(X le 6) = frac66 = 1$$
Also:
$$P(X le 7) = frac66 = 1$$
$$P(X le 0) = frac06$$
$$P(X le 0.5) = frac06$$
$$P(X le -2000) = frac06$$
$$P(X le 4.5) = frac46$$
$$P(X > 7) = frac06$$
Thus:
The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'
The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'
Also
Show that $Y_n$ converges in probability
edited Apr 13 '17 at 12:19
Community♦
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
$endgroup$
add a comment |
$begingroup$
Imagine we have $X$ representing the toss of a fair die.
Then $text{Range}(X) = {1,2,cdots,6}$
$$P(X = x) = frac16$$
So:
$$P(X le 1) = frac16$$
$$P(X le 2) = frac26$$
$$P(X le 3) = frac36$$
$$P(X le 4) = frac46$$
$$P(X le 5) = frac56$$
$$P(X le 6) = frac66 = 1$$
Also:
$$P(X le 7) = frac66 = 1$$
$$P(X le 0) = frac06$$
$$P(X le 0.5) = frac06$$
$$P(X le -2000) = frac06$$
$$P(X le 4.5) = frac46$$
$$P(X > 7) = frac06$$
Thus:
The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'
The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'
Also
Show that $Y_n$ converges in probability
edited Apr 13 '17 at 12:19
Community♦
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
$endgroup$
Imagine we have $X$ representing the toss of a fair die.
Then $text{Range}(X) = {1,2,cdots,6}$
$$P(X = x) = frac16$$
So:
$$P(X le 1) = frac16$$
$$P(X le 2) = frac26$$
$$P(X le 3) = frac36$$
$$P(X le 4) = frac46$$
$$P(X le 5) = frac56$$
$$P(X le 6) = frac66 = 1$$
Also:
$$P(X le 7) = frac66 = 1$$
$$P(X le 0) = frac06$$
$$P(X le 0.5) = frac06$$
$$P(X le -2000) = frac06$$
$$P(X le 4.5) = frac46$$
$$P(X > 7) = frac06$$
Thus:
The '$varepsilon > theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$X > 7$'
The '$0 < varepsilon < theta$' (I'll assume '$varepsilon > 0$' is a typo) is like '$1 < X < 6$'
Also
Show that $Y_n$ converges in probability
edited Apr 13 '17 at 12:19
Community♦
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
answered Nov 2 '16 at 15:04
BCLCBCLC
1
1
add a comment |
add a comment |
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