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Conditional Distributions and Rationals
Conditional expectation of $X$ given $sin(X)$Uniform measure on the rationals between 0 and 1Must every probability distribution over a countable set be discrete?Conditional expectation of the maximum of two independent uniform random variables given one of themConditional probability integral transformConditional expectation of the maximum of two independent random variables, given one of themAre conditional probabilities only work on uniform distribution?Illustration of conditional distributionProbability - Calculate E[y] given that you know the conditional density of Y given X = xConditional distributions, finding the marginal pdf
$begingroup$
If $U$ is uniformly distributed on $[0, 1]$, what is the conditional distribution of $U$ given that $U$ is rational? Intuitively, it would be uniform, but we cannot have a uniform distribution over a countable set.
real-analysis probability-theory measure-theory
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
$endgroup$
add a comment |
$begingroup$
If $U$ is uniformly distributed on $[0, 1]$, what is the conditional distribution of $U$ given that $U$ is rational? Intuitively, it would be uniform, but we cannot have a uniform distribution over a countable set.
real-analysis probability-theory measure-theory
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
$endgroup$
$begingroup$
"$U$ is rational" is a measure zero event, so the sigma algebra generated by $U$ has only events of measure zero or one. Therefore, the conditional expectation is just the constant random variable given by the expectation of $U$, which is half. (If you are speaking of conditional expectation over a sigma algebra as a random variable).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 19 at 3:32
add a comment |
$begingroup$
If $U$ is uniformly distributed on $[0, 1]$, what is the conditional distribution of $U$ given that $U$ is rational? Intuitively, it would be uniform, but we cannot have a uniform distribution over a countable set.
real-analysis probability-theory measure-theory
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
$endgroup$
If $U$ is uniformly distributed on $[0, 1]$, what is the conditional distribution of $U$ given that $U$ is rational? Intuitively, it would be uniform, but we cannot have a uniform distribution over a countable set.
real-analysis probability-theory measure-theory
real-analysis probability-theory measure-theory
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
asked Mar 19 at 3:28
Kurtland ChuaKurtland Chua
3541212
3541212
$begingroup$
"$U$ is rational" is a measure zero event, so the sigma algebra generated by $U$ has only events of measure zero or one. Therefore, the conditional expectation is just the constant random variable given by the expectation of $U$, which is half. (If you are speaking of conditional expectation over a sigma algebra as a random variable).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 19 at 3:32
add a comment |
$begingroup$
"$U$ is rational" is a measure zero event, so the sigma algebra generated by $U$ has only events of measure zero or one. Therefore, the conditional expectation is just the constant random variable given by the expectation of $U$, which is half. (If you are speaking of conditional expectation over a sigma algebra as a random variable).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 19 at 3:32
$begingroup$
"$U$ is rational" is a measure zero event, so the sigma algebra generated by $U$ has only events of measure zero or one. Therefore, the conditional expectation is just the constant random variable given by the expectation of $U$, which is half. (If you are speaking of conditional expectation over a sigma algebra as a random variable).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 19 at 3:32
$begingroup$
"$U$ is rational" is a measure zero event, so the sigma algebra generated by $U$ has only events of measure zero or one. Therefore, the conditional expectation is just the constant random variable given by the expectation of $U$, which is half. (If you are speaking of conditional expectation over a sigma algebra as a random variable).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 19 at 3:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The rationals form a null set, i.e. a set of probability $0$, and conditional probabilities given a null set are not defined.
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The rationals form a null set, i.e. a set of probability $0$, and conditional probabilities given a null set are not defined.
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
add a comment |
$begingroup$
The rationals form a null set, i.e. a set of probability $0$, and conditional probabilities given a null set are not defined.
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
add a comment |
$begingroup$
The rationals form a null set, i.e. a set of probability $0$, and conditional probabilities given a null set are not defined.
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
$endgroup$
The rationals form a null set, i.e. a set of probability $0$, and conditional probabilities given a null set are not defined.
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
answered Mar 19 at 3:32
Robert IsraelRobert Israel
330k23219473
330k23219473
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
add a comment |
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
Why are there cases where we do such conditioning? For example, say $(X_n)$ is a Poisson process - then, for a fixed $n$, we can talk about the joint conditional distribution of $X_1, dots, X_n$ given $X_{n+1} = x$, even if the event that $X_{n+1} = x$ has measure zero.
$endgroup$
– Kurtland Chua
Mar 19 at 3:41
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@KurtlandChua : What do you mean by "$(X_n)$ is a Poisson process"? Poisson processes $N(t)$ take integer values and $P[N(t)=k]>0$ for all $k$. You might want another example like "what if $X$ is uniform over $[0,1]$" for which things like $E[Y|X=x]$ or $E[Y|sigma(X)]$ make sense. (I would be interested if Robert has any insight/discussion of the distinctions.)
$endgroup$
– Michael
Mar 19 at 4:16
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
@Michael Oh, I guess going off of your definition, what I had in mind was $X_n = sup_{t > 0, N(t) < n} t$ (i.e. when the arrivals happen)
$endgroup$
– Kurtland Chua
Mar 19 at 4:39
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
$begingroup$
There is an exception for conditioning with respect to the value of a continuous random variable. So you can talk about conditional probabilities given $U = u$ for $u in [0,1]$. But not conditional probabilities given $U in $ some set.
$endgroup$
– Robert Israel
Mar 19 at 14:12
add a comment |
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$begingroup$
"$U$ is rational" is a measure zero event, so the sigma algebra generated by $U$ has only events of measure zero or one. Therefore, the conditional expectation is just the constant random variable given by the expectation of $U$, which is half. (If you are speaking of conditional expectation over a sigma algebra as a random variable).
$endgroup$
– астон вілла олоф мэллбэрг
Mar 19 at 3:32