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This seems almost silly to ask but I am stuck with it.

I have the following equation

\begin{equation}\tag{*} e^{-\frac{(A-B)^{2}}{D}}=C \end{equation}

I know $A,B \in \mathbb{R}, 0<C<1,D>0 $. Everything is known except $A$. I computed $A$ in following way.

\begin{alignedat}{2} \Rightarrow\quad && {\frac{(A-B)^{2}}{D}}=-\ln C \\ \Rightarrow\quad && {{(A-B)^{2}}}=D*\ln \big(\frac{1}{C}\big) \\ \Rightarrow\quad && {{(A-B)}}=\sqrt {D*\ln \big(\frac{1}{C}\big)} \\ \Rightarrow\quad && {A}= B + \sqrt {D*\ln \big(\frac{1}{C}\big)} \end{alignedat}

But when I use $A$ in (*) along with $B,D$, I get a value of $C$ that is different from what I originally had. Any hint/suggestion will be much appreciated.

NAASI
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  • Please list the values you have, so we can check your work. It is likely to be the sign you gave the square root. In some cases it might be numeric inaccuracy. – Ross Millikan Mar 20 '15 at 01:21

1 Answers1

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Your method does not work for all $A, B, C, N \in \mathbb{R}$.

  • What is $\ln C$, if $C \le 0$?
  • What is $\sqrt{N \ln \frac{1}{C}}$, if $N \ln \frac{1}{C} \lt 0$ (ie. if $N \lt 0$ and $C \gt 1$ or $N \gt 0$ and $0 \lt C \lt 1$)
  • $x^2 = y$ has two solutions $x$ for positive $y$. What about the other solution, here namely $-\sqrt{N \ln \frac{1}{C}}$?

Depending on your values of $B, C$ and $N$ in one of these steps you made your error. (I assume you just forgot the root sign in your last line?) As you get a real result, I conclude you forgot the other solution of your second line (point 3 here), but you approach could have gone wrong even earlier.

Keba
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  • We should be guaranteed that $C \gt 0$ so we can take the log. Otherwise there is no real solution. I would bet on the sign of the square root. – Ross Millikan Mar 20 '15 at 01:19
  • I totally agree. I would also assume $N \in \mathbb{N}$, since else it would be named $D$ instead. ;) – Keba Mar 20 '15 at 01:24
  • Thank you for your comments. I ll make an edit to improve my question – NAASI Mar 20 '15 at 01:44
  • To put this into perspective, the given expression is a Normal distribution. A is my noisy observation, B is noiseless observation and noise normally distributed with mean=0 and var=sigma^2. D is 2*sigma^2. This is expression is common expression in Maximum Likelihood/ Max A-posteriori Detection. – NAASI Mar 20 '15 at 02:05
  • You still forget about $A = B - \sqrt{\dots}$ – Keba Mar 20 '15 at 02:12
  • Since C is a measure of probability therefore 0<C<1, N or D is variance therefore its in always >=0. This would mean that ln(1/C)>0. sqrt(N*ln(1/C)) will be always real. Now its the sign od sqrt(.) that needs to be sorted – NAASI Mar 20 '15 at 02:25