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In my textbook it says that the probability of obtaining a value (which its Gauss distribution is known) is simply $$\frac{1}{\sqrt{2\pi }} \cdot e^{\frac{(x_{1}-X)^{2}}{2\sigma ^{2}}} \text{d}x$$ where $x_{1}$ is the value for which we are seeking its probability. The book then says that we can drop the $\text{d}x$ and it will not affect the result. However, since the normal distribution is normalized and the area under the curve represent the probability, it seems like dropping the $\text{dx}$ will no longer give us a probability. Could someone explain if dropping the $\text{d}x$ will affect the result of the probability of a single value?

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I believe that when you are talking of probabilities in this case, you do so thinking of a continuous domain of values for $x$. If so, the expression you have is rather the product of a probability density function (p.d.f) by $dx$ than the probability of finding $x_{1}$ itself.
According to David J. Griffiths, in his Introduction to Quantum Mechanics textbook, there is often a misconception about the meaning of the p.d.f, that represents, when integrated in an interval $[a, b]$ in relation to $x$, the probability of obtaining the desired value (in this case, $x_{1}$) within such an interval.
So, as you thought, dropping the $dx$ will no longer represent a probability, because it never did! Indeed, if you calculate $$\int_{a}^{b} \rho(x) dx$$ for $a = b = x_{1}$, the integral will give you zero, that is precisely the probability of finding $x_{1}$ on a finite, but continuous interval.

UPDATE
Accordingly to the passage in the book that you talked about in the comments (An Introduction to Error Analysis), we can, indeed, drop the $dx$ and still maintain the probability of getting $x_1$. But here, as the author himself says, we are no longer strictly talking about the probability of having $x_1$ measured, but of measuring a value in the vincinity of $x_1$ (that's why he drops the $dx$ and the $\frac{1}{\sqrt{2\pi}}$, because they are not really important in the reasoning of the section).
So, in the end what the author is doing is to omit the factors $dx$ because they would add an inconvenient term to the product of probabilities. The following paragraphs will, hopefully, clarify that.

$$Prob(x \in \mathbb{R}\ | x_1 \leq x \leq x_1 +dx_1) = \frac{1}{\sigma\sqrt{2\pi}}\exp^{-(x_1 - X)^2/2\sigma^2}dx_1 \tag{1}$$ In practice, we are not interested in the size of the interval $dx_1$ (or the factor $\sqrt{2\pi}$), so we abbreviate this equation to $$Prob(x_1) \propto \frac{1}{\sigma}\exp^{-(x_1 - X)^2/2\sigma^2}. \tag{2}$$

As you can see, the author replaces the equality symbol by a proporcionality one, what means that the second equation says that the probability of finding $x$ in the vincinity of $x_1$ is proportional to the expression, not precisely equal. In the final of the section, he uses the product of all of these "approximations" to calculate the $Prob(x_1, x_2, ..., x_N)$, what would be, in what concerns the notation, really bad to write without surppressing the $dx_k$ as you can see below $$Prob_{X,\sigma}(x_1, x_2, ..., x_N) = \frac{1}{\sigma^{N}(\sqrt{2\pi})^{N}}\exp^{-\sum(x_i - X)^2/2\sigma^2}\prod\limits_{k = 1}^N dx_k. \tag{3}$$ $$Prob_{X, \sigma}(x_1, x_2, ..., x_N) \propto \frac{1}{\sigma^N}\exp^{-\sum(x_i - X)^2/2\sigma^2}. \tag{4}$$ Finally, as one of the goals of the section is to find the maximum value of the expressions above, due to $X$ and $\sigma$, there is no difference between using $(3)$ or $(4)$, except that the later is compacter.

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  • $\begingroup$ This is exactly what I thought, the probability of finding x1 is zero. However, even the probability of finding any value in the interval [x1, x+dx] is still can't be expressed with the dropping the dx. However the book says that we can still get the same probability when dropping the dx. The book is Taylor Introduction to Error Analysis page 138. Also in page 234. $\endgroup$ – NegativeTension Sep 28 '17 at 18:58
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The probability density function helps you decide the probability of observing a value in a particular range - only the integral form makes sense when you speak of "probability" (as opposed to "probability density"). The probability of a single value is zero - so if you want the probability to get a specific value, you should not only drop the dx, but the rest of the expression as well.

On the other hand - if you say "getting a value" means "getting a value within a certain precision which is determined by the resolution of my instrument, R" then you can replace $dx$ with $R$ and thus obtain the probability.

Of course, if you are only interested in the relative probabilities of obtaining one value vs another, and again you have some finite rounding in your measurement system, then dropping the $dx$ term appears legitimate since the terms would cancel.

I wonder if you should look for another text book.

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