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Short question

Is there an equivalent to the proportionality sign $\propto$ for additive constants? The proportionality relation $y\propto x$ implies that $y=kx$ for some constant $k$. Is there a shorthand to express that $y=x+k$ for some constant $k$?

Long question

Probability distributions are often used without their explicit normalisations. E.g. for a normal distribution may be written as $$ P(x)\propto \exp\left(-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}\right). $$

When working with log-probabilities it would be convenient to have a relation expressing that two quantities are equal up to a normalising constant such that $$ \log P(x)\square-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}, $$ where $\square$ is the desired relation.

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  • $\begingroup$ I don't know of a general mathematical equivalent. There may be an accepted one in statistics for the case you really care about. Delete hear and ask on stats.stackexchange.com? $\endgroup$ Nov 1, 2017 at 17:02
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    $\begingroup$ I'm not aware of such a symbol, though your question did remind me of what happens when computing indefinite integrals, where the answer is only determined up to an additive constant. In that case everyone is used to seeing a '$+ C$' pop up, and you could use the same notation here. Though I'd probably add the comment 'for some constant $C$' the first few times I did it. $\endgroup$ Nov 1, 2017 at 17:18

2 Answers 2

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You can simply define a relation $\sim$ between functions such that

$$f \sim g \Longleftrightarrow \exists k \in \mathbb{R} . f = g + k$$

Or (slightly) more formally,

$$f \sim g \Longleftrightarrow \exists k \in \mathbb{R} . \forall x \in X . f(x) = g(x) + k$$

where $X$ is the domain of your functions.

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Proportional and linear functions are almost identical in form. The only difference is the addition of the “b” constant to the linear function. Indeed, a proportional relationship is just a linear relationship where b = 0, or to put it another way, where the line passes through the origin (0,0). So in fact, a proportional relationship is just a special kind of linear relationship, i.e., all proportional relationships are linear relationships (although not all linear relationships are proportional).

Your case comes under Not all linear are Proportional.

Based on whether it is monotonically decreasing or increasing the proportionality might be direct or indirect.

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  • $\begingroup$ You point out that not all linear relationships are proportional. For that reason you shouldn't use the proportional symbol. $\endgroup$
    – Mark S.
    Nov 1, 2017 at 16:30
  • $\begingroup$ @MarkS. Ya i agree with you. I removed the proportionality $\endgroup$
    – loneStar
    Nov 1, 2017 at 16:58

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