Can Iverson brackets be thought of as returning probabilities?
$begingroup$
An Iverson bracket provides a way to take a formula with a truth value and translate it back into a number. This can simplify some things like summation bounds and integral bounds.
Here an upper bound of $100$ is written with an Iverson bracket, the expression $[i le 100]$ evaluates to $1$ if $i$ is less than or equal to $100$ and $0$ otherwise.
$$ sum_{i=0}^{100} f(n) = sum_{i in mathbb{N}}[i le 100]f(n) tag{1} $$
I can think of another circumstance where it's common to convert something resembling a proposition to a number, the probability of operator $mathbb{P}[dots]$ , e.g.
$$ mathbb{P}[lnot psi] = 1-mathbb{P}[psi] tag{2} $$
Is it right to think of $mathbb{P}[dots]$ as a conservative extension of $[dots]$? In cases where the random variables are not involved in the argument to the bracket, the probability must be $0$ or $1$ . This makes it easy to do nifty things like define the pdf in terms of a random variable:
$$ mathrm{pdf}_Y(s) stackrel{df}{=} frac{mathrm{d},[Y le s]}{mathrm{d}s} tag{3} $$
Are there any pitfalls or hidden subtleties to thinking of an Iverson bracket this way / extending the notion of an Iverson bracket to expressions with probabilities rather than crisp truth values?
notation
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
$endgroup$
add a comment |
$begingroup$
An Iverson bracket provides a way to take a formula with a truth value and translate it back into a number. This can simplify some things like summation bounds and integral bounds.
Here an upper bound of $100$ is written with an Iverson bracket, the expression $[i le 100]$ evaluates to $1$ if $i$ is less than or equal to $100$ and $0$ otherwise.
$$ sum_{i=0}^{100} f(n) = sum_{i in mathbb{N}}[i le 100]f(n) tag{1} $$
I can think of another circumstance where it's common to convert something resembling a proposition to a number, the probability of operator $mathbb{P}[dots]$ , e.g.
$$ mathbb{P}[lnot psi] = 1-mathbb{P}[psi] tag{2} $$
Is it right to think of $mathbb{P}[dots]$ as a conservative extension of $[dots]$? In cases where the random variables are not involved in the argument to the bracket, the probability must be $0$ or $1$ . This makes it easy to do nifty things like define the pdf in terms of a random variable:
$$ mathrm{pdf}_Y(s) stackrel{df}{=} frac{mathrm{d},[Y le s]}{mathrm{d}s} tag{3} $$
Are there any pitfalls or hidden subtleties to thinking of an Iverson bracket this way / extending the notion of an Iverson bracket to expressions with probabilities rather than crisp truth values?
notation
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
$endgroup$
add a comment |
$begingroup$
An Iverson bracket provides a way to take a formula with a truth value and translate it back into a number. This can simplify some things like summation bounds and integral bounds.
Here an upper bound of $100$ is written with an Iverson bracket, the expression $[i le 100]$ evaluates to $1$ if $i$ is less than or equal to $100$ and $0$ otherwise.
$$ sum_{i=0}^{100} f(n) = sum_{i in mathbb{N}}[i le 100]f(n) tag{1} $$
I can think of another circumstance where it's common to convert something resembling a proposition to a number, the probability of operator $mathbb{P}[dots]$ , e.g.
$$ mathbb{P}[lnot psi] = 1-mathbb{P}[psi] tag{2} $$
Is it right to think of $mathbb{P}[dots]$ as a conservative extension of $[dots]$? In cases where the random variables are not involved in the argument to the bracket, the probability must be $0$ or $1$ . This makes it easy to do nifty things like define the pdf in terms of a random variable:
$$ mathrm{pdf}_Y(s) stackrel{df}{=} frac{mathrm{d},[Y le s]}{mathrm{d}s} tag{3} $$
Are there any pitfalls or hidden subtleties to thinking of an Iverson bracket this way / extending the notion of an Iverson bracket to expressions with probabilities rather than crisp truth values?
notation
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
$endgroup$
An Iverson bracket provides a way to take a formula with a truth value and translate it back into a number. This can simplify some things like summation bounds and integral bounds.
Here an upper bound of $100$ is written with an Iverson bracket, the expression $[i le 100]$ evaluates to $1$ if $i$ is less than or equal to $100$ and $0$ otherwise.
$$ sum_{i=0}^{100} f(n) = sum_{i in mathbb{N}}[i le 100]f(n) tag{1} $$
I can think of another circumstance where it's common to convert something resembling a proposition to a number, the probability of operator $mathbb{P}[dots]$ , e.g.
$$ mathbb{P}[lnot psi] = 1-mathbb{P}[psi] tag{2} $$
Is it right to think of $mathbb{P}[dots]$ as a conservative extension of $[dots]$? In cases where the random variables are not involved in the argument to the bracket, the probability must be $0$ or $1$ . This makes it easy to do nifty things like define the pdf in terms of a random variable:
$$ mathrm{pdf}_Y(s) stackrel{df}{=} frac{mathrm{d},[Y le s]}{mathrm{d}s} tag{3} $$
Are there any pitfalls or hidden subtleties to thinking of an Iverson bracket this way / extending the notion of an Iverson bracket to expressions with probabilities rather than crisp truth values?
notation
notation
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
asked Dec 25 '18 at 0:38
Gregory NisbetGregory Nisbet
734612
734612
add a comment |
add a comment |
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