# Is this sensible: $P(y_{1}<Y\le y_{2}\mid Y\sim\mathcal{D}(\mu))$

I want to write:

$$P(y_{1}<Y\le y_{2}\mid Y\sim\mathcal{D}(\mu))$$

to say:

• The probability of $Y$ being between $y_1$ and $y_2$
• given that $Y$ is a random variable distributed according to distribution $\mathcal{D}$
• when $\mathcal{D}$ is parameter by $\mu$

I want to leave the $\mathcal{D}$ in because $\mathcal{D}$ could be one of a number of different distributions, all of which are parameterised the same way.

but I've never seen a "distributed as" ($\sim$) on the right hand side of "given that" ($\mid$)

I was considering other notations:

• $$P_{Y\sim\mathcal{D}(\mu)}(y_{1}<Y\le y_{2}\mid \mu)$$
• $$P(y_{1}<\mathcal{D}(\mu)\le y_{2})$$
• $$F_{\mathcal{D}(\mu)}(y_2) - F_{\mathcal{D}(\mu)}(y_1)$$

I want to use this as part of a numerical procedure to describe a way that a continuous space point is mapped to a probability mass vector: E.g. something like a function $\Omega$

$$\Omega: \mathbb{R}\times\mathbb{D} \to \mathbb{R}^{100}$$ For $\mathbb{D}$ the space of 1-D continuous distributions parameterised by 1 scalar $$\Omega(v,\mathcal{D})=\left(P\left(\dfrac{i-1}{100}<Y\le\dfrac{i}{100}\mid Y\sim\mathcal{D}(v)\right)\right)_{i=1}^{i=100}$$

So this is basically mapping a point $v$ and distribution $\mathcal{D}$ to histograms of that distribution when the first paramater is given by $v$

The random variable $Y$ is itself without real meaning -- just a step in describing the process

• It's a slight abuse of notation, because you are not actually conditioning on anything, but people will understand easily what you mean. – Gordon Smyth Jun 30 '17 at 4:47
• I deleted my post...your description of $\Omega(v,\mathcal{D})$ was pretty clear. However, where did $10$ come from in the second parameter? – user145807 Jun 30 '17 at 5:13
• In this case it is an arbitrary value for the example, I'll just remove it entirely. – Lyndon White Jun 30 '17 at 5:15
• I'd say if you use your "conditional" notation after explaining your mapping, then I'd think I could follow the rest of your argument...but by itself it's not self explanatory. – user145807 Jun 30 '17 at 5:28

This is an abuse of notation. How would you calculate $P(\,Y\sim\mathcal{D}(\mu)\,)$ ? $Y\sim\mathcal{D}$ is a definition that states: $Y$ is a random variable that is distributed as $\mathcal{D}$. If you say that $Y$ can follow different probability distributions, then what is the probability distribution of $Y$? For it to be a random variable it needs to have a probability distribution, so it cannot have multiple probability distributions.
• What notation do you suggest instead? Your example of $P(Y \sim D)$ makes it clear that using this notation conflicts with considering it as a likihood function – Lyndon White Jun 30 '17 at 8:18
• @LyndonWhite but for the reasons outlined above: $Y$ cannot follow multiple probability distributions at the same time! – Tim Jun 30 '17 at 8:39
• At the same time it can not. But the function $\Omega$ in my example is a reasonable function from the space of pairs of reals and distributions to the space of n-vectors of reals. – Lyndon White Jun 30 '17 at 8:51
• Thing is $Y$ is not a random variable that exists outside of the definition of the function. It is not the distribution of something in a population. It is I guess a way of expressing evaluating CDFs in various sections – Lyndon White Jun 30 '17 at 9:02