Motivational example for introducing singular distributions fom Rosenthal
$begingroup$
I am studying Rosenthal´s book, and I am stuck on one example, where author is showing why it is neccesary to consider distributions diffrent from absolutely continuous or discrete.
Well, what I don´t understand is the fact, that $P(Y in S)=1$ (by law of large numbers, I dont see how the average of partial sums converges to 2/3) and also why the measure of introduced set is $lambda = 0.$
Here is the screen of the motivational example.
Thanks in advance for any help.
probability-theory measure-theory random-variables lebesgue-measure
asked Dec 14 '18 at 15:44
stanlystanly
699
$endgroup$
add a comment |
$begingroup$
I am studying Rosenthal´s book, and I am stuck on one example, where author is showing why it is neccesary to consider distributions diffrent from absolutely continuous or discrete.
Well, what I don´t understand is the fact, that $P(Y in S)=1$ (by law of large numbers, I dont see how the average of partial sums converges to 2/3) and also why the measure of introduced set is $lambda = 0.$
Here is the screen of the motivational example.
Thanks in advance for any help.
probability-theory measure-theory random-variables lebesgue-measure
asked Dec 14 '18 at 15:44
stanlystanly
699
$endgroup$
add a comment |
$begingroup$
I am studying Rosenthal´s book, and I am stuck on one example, where author is showing why it is neccesary to consider distributions diffrent from absolutely continuous or discrete.
Well, what I don´t understand is the fact, that $P(Y in S)=1$ (by law of large numbers, I dont see how the average of partial sums converges to 2/3) and also why the measure of introduced set is $lambda = 0.$
Here is the screen of the motivational example.
Thanks in advance for any help.
probability-theory measure-theory random-variables lebesgue-measure
asked Dec 14 '18 at 15:44
stanlystanly
699
$endgroup$
I am studying Rosenthal´s book, and I am stuck on one example, where author is showing why it is neccesary to consider distributions diffrent from absolutely continuous or discrete.
Well, what I don´t understand is the fact, that $P(Y in S)=1$ (by law of large numbers, I dont see how the average of partial sums converges to 2/3) and also why the measure of introduced set is $lambda = 0.$
Here is the screen of the motivational example.
Thanks in advance for any help.
probability-theory measure-theory random-variables lebesgue-measure
probability-theory measure-theory random-variables lebesgue-measure
asked Dec 14 '18 at 15:44
stanlystanly
699
asked Dec 14 '18 at 15:44
stanlystanly
699
asked Dec 14 '18 at 15:44
stanlystanly
699
asked Dec 14 '18 at 15:44
stanlystanly
699
asked Dec 14 '18 at 15:44
stanlystanly
699
699
add a comment |
add a comment |
1 Answer
1
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oldest
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Since the $Z_n$ are i.i.d, by the strong-law of large numbers
$$
frac{d_1(Y)+dotsb d_n(Y)}{n}=frac{Z_1+dotsb+Z_n}{n}to EZ_1=P(Z_1=1)=frac{2}{3}
$$
almost surely as $nto infty$. Hence $P(Yin S)=1.$
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
$endgroup$
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since the $Z_n$ are i.i.d, by the strong-law of large numbers
$$
frac{d_1(Y)+dotsb d_n(Y)}{n}=frac{Z_1+dotsb+Z_n}{n}to EZ_1=P(Z_1=1)=frac{2}{3}
$$
almost surely as $nto infty$. Hence $P(Yin S)=1.$
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
$endgroup$
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
add a comment |
$begingroup$
Since the $Z_n$ are i.i.d, by the strong-law of large numbers
$$
frac{d_1(Y)+dotsb d_n(Y)}{n}=frac{Z_1+dotsb+Z_n}{n}to EZ_1=P(Z_1=1)=frac{2}{3}
$$
almost surely as $nto infty$. Hence $P(Yin S)=1.$
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
$endgroup$
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
add a comment |
$begingroup$
Since the $Z_n$ are i.i.d, by the strong-law of large numbers
$$
frac{d_1(Y)+dotsb d_n(Y)}{n}=frac{Z_1+dotsb+Z_n}{n}to EZ_1=P(Z_1=1)=frac{2}{3}
$$
almost surely as $nto infty$. Hence $P(Yin S)=1.$
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
$endgroup$
Since the $Z_n$ are i.i.d, by the strong-law of large numbers
$$
frac{d_1(Y)+dotsb d_n(Y)}{n}=frac{Z_1+dotsb+Z_n}{n}to EZ_1=P(Z_1=1)=frac{2}{3}
$$
almost surely as $nto infty$. Hence $P(Yin S)=1.$
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
edited Dec 14 '18 at 16:17
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
answered Dec 14 '18 at 15:52
Foobaz JohnFoobaz John
22.4k41452
22.4k41452
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
add a comment |
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
Oh, I see. I didn't notice, that $d_{i}(Y)$ is actually $Z_{i}.$ But, I would have second question: why is the mesure of $S$ equal to $0.$ What´s the reason, that it contains only countably many numbers?
$endgroup$
– stanly
Dec 14 '18 at 16:01
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
$begingroup$
No, it's uncountable. But almost all members of $[0,1]$ have that limit $= 1/2$ rather than $2/3$, again by the strong law of large numbers.
$endgroup$
– Robert Israel
Dec 14 '18 at 16:32
add a comment |
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