How can I divide $\frac{\frac{1}{8}\delta (x-0,y-0) + \frac{1}{4}\delta (x-2,y-0) + \frac{3}{8}\delta(x-0, y-4) + \frac{1}{4}\delta (x-2,y-4)}{\frac{1}{2}\delta(x-0) + \frac{1}{2}\delta (x-2)}$?
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1How does this come up? Generally "dividing by the Dirac delta function" simply doesn't make sense. – Matt Samuel Feb 21 '15 at 21:15
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I was doing a problem involving conditional probability. ^ This isnt the only way to solve the problem, but I was just curious. – user3642365 Feb 21 '15 at 21:16
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1I recommend pursuing other avenues because this way won't work. – Matt Samuel Feb 21 '15 at 21:17
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In the numerator is written as: $\frac{1}{8}\delta(x)(\delta(y)+3\delta(y-4))+\frac{1}{4}\delta(x-2)(\delta(y)+\delta(y-4))$. You use $\delta(x,y)=\delta(x)\delta(y)$ and simplify.
Now you can use the relation: $\frac{\delta(x)}{\delta(x)+\delta(x-2)}$ is zero for $x=2$ (because you divide by $\infty$ at this point) and 1 for $x \neq 0$ due to L'Hospital rule.
$\frac{\delta(x-2)}{\delta(x)+\delta(x-2)}$ is zero on $x=0$ and 1 elsewhere.
kryomaxim
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L'Hospital's rule? That involves taking derivatives. What's the derivative of the Dirac delta function? – Matt Samuel Feb 21 '15 at 21:26
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2Actually, delta has a derivative, but everything else doesn't make sense (and delta is not a classical function, it's a measure or a distribution). And, @jm324354, unit step function is not the derivative of delta, it's delta the derivative of the unit step function (in a sense where talking about a derivative of the unit step function makes sense) – Tryss Feb 21 '15 at 21:36
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1Backwards. According to physicists the derivative of the unit step function is the Dirac delta function. – Matt Samuel Feb 21 '15 at 21:36
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The explaination with L'Hospital rule was confusing, sorry! I meant that $\frac{\infty}{\infty}=1$. – kryomaxim Feb 21 '15 at 21:37