I have the following equation that I would like to solve for $y$: $$\alpha=\sum_{i=1}^{n}\exp(-y\beta_i)$$ $\alpha$, $y$ and $\beta_i$ are all positive reals (the $\beta_i$'s are independent random numbers), and $n$ is a large positive integer. I'm unable to find a way to isolate $y$. Is this equation solvable?

Probability distribution of $\beta_i$:

$\beta_i$ is defined as $\beta_i=\sqrt{2\omega_i}$, where $\omega_i$ assumes a uniform probability distribution in some interval $[\omega_a,\omega_b]$, where $0\leq\omega_a<\omega_b$.

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    $\begingroup$ No, in general this equation is not solvable for y. Espacially when $n$ is a large number. $\endgroup$ – Cornman Jun 17 '19 at 13:07
  • $\begingroup$ As @Cornman said, you won't be able to get rid of that sum of exponentials $\endgroup$ – David Jun 17 '19 at 13:18
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    $\begingroup$ If $\beta_i$ were all rational numbers, then this will become equivalent to solving a polynomial of a potentially enormous degree (probably much larger than $n$). Naturally, there is no formula for solving polynomials of degree $5$ or more, so you might be out of luck. $\endgroup$ – Theo Bendit Jun 17 '19 at 13:20
  • $\begingroup$ hmm.. ok, thank you $\endgroup$ – Tofi Jun 17 '19 at 13:25
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    $\begingroup$ It is worth noting that there is a unique real solution $y$ for any values of $\alpha$ and the $\beta_i$. Given $\alpha$ and the $\beta_i$ you can approximate $y$ with the help of a computer. An explicit expression for $y$ in terms of $\alpha$ and the $\beta_i$ is too much to ask, as the other comments explain. $\endgroup$ – Servaes Jun 17 '19 at 13:29

Updated post

After publishing my original post the author of the OP has specified the PDF for the $\beta_i$, also hereby he has assumed that all $\beta_i$ have the same distribution (which was my main simplifying assumption).

Here I take up this definition and update my solution.

EDIT (18.06.19): the following through derivation yields the corrected result which is even simpler than the previous one.

The distribution function for $b$ is from the definition $w = \frac{1}{2}b^2$ and $f(w)$ flat between $w_a$ and $w_b$ given from $dw = b\; db$ after normalization

$$f(b) = \frac{2 b}{b_b^2-b_a^2} = \frac{b}{w_b-w_a}$$

Now we can do the integral $(2)$ which gives

$$\alpha = n \int_{b_a}^{b_b} f(b) e^{- y b} \, db \\ = \frac{2 n \left(e^{- y b_a } (y b_a+1)-e^{-y b_b} (y b_b +1)\right)}{y^2 \left(b_b^2-b_a^2\right)}\tag{4a} $$

We have obtained a transcendental equation for $y(\alpha)$. This is best solved in the inverted form $(4a)$ for $\alpha (y)$.

In the limit of a very narrow distribution ($w_a \to w_b (=w)$) we get from $(4a)$

$$\alpha = n e^{-b_b y}\tag{4b} $$

which can be inverted to give the explicit solution

$$y = \frac{1}{b_b}\log(\frac{\alpha}{n})\tag{4c}$$

My original post

Making the simplifying assumptions that the $\beta_i$ have the same PDF $f(\beta)$ we can take the expectation value of the equation over all $\beta_i$ and get

$$E(\alpha) = \alpha = \sum_{i=1}^n E(\exp(- y \beta_i))= n E(\exp(- y \beta))\tag{1}$$

More explicitly we have for a continuous variable $\beta$

$$E(\exp(- y \beta))=\int_{0}^\infty f(\beta) \exp(- y \beta))^\,d\beta\tag{2}$$

Assuming for simplicity that $\beta$ is exponentially distributed, i.e.

$$f(\beta) = a \exp(- a \beta )\tag{3}$$

the integral can be done, and we obtain

$$\alpha = n \frac{a}{a + y}\tag{4}$$

which can be solved explicitly for $y$ to give

$$y = a (\frac{n}{\alpha}-1)\tag{5}$$

  • $\begingroup$ Thank you, it looks like this equation is just too much! $\endgroup$ – Tofi Jun 17 '19 at 23:42
  • $\begingroup$ @ Tofi Why "too much"? The specific problem is solved by equation $(4a)$ leaving just the standard problem of solving numerically a transcendental equation. $\endgroup$ – Dr. Wolfgang Hintze Jun 18 '19 at 8:06
  • $\begingroup$ Yeah, but what I wanted was a nice analytical expression. By the way, the value I found for the PDF of $\beta$ is $\dfrac{\beta}{\omega_b-\omega_a}$, so is this the same as yours? $\endgroup$ – Tofi Jun 18 '19 at 11:36

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