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The variable t is defined in the domain $t>0$. At first, I thought $k$ could be calculated by normalizing $p(t)$ to $1$. So $$k = \frac{1}{F(\infty) - F(0)}$$ where $F(t)$ is the primitive function of $f(t)$.

However, I encountered a problem with this type of function, (I applied specific values to give a concrete example):

\begin{equation} p(t) = k(1+\sin(20t))e^{\frac{-t}{1.5}} \end{equation}

when using the method above, I found $$F(\infty) - F(0) \approx 1.55$$ But, when plotted, $f(t)$ looks like this:

enter image description here

So for some $t\to 0$, where $f(t)$ peaks, the probability $p(t)$ will be larger than $1$ because $1/k$ is smaller than $f(t)$ there. In this case, how do I find $k$? What constraints should I put in my calculation to make sure that $p(t)$ is always smaller than $1$.

Sebastiano
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Winniebear
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  • Only the cumulative distribution function has to be smaller than one. The probability density function may be strictly larger than one. See also here: https://math.stackexchange.com/questions/105455/how-can-a-probability-density-be-greater-than-one-and-integrate-to-one. (Of course, it may only be strictly larger than one on a sufficiently small interval.) –  Dec 28 '20 at 21:28

1 Answers1

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It's the integral of density over its domain that needs to be $1,$ not the density itself, which in some cases can exceed $1$ at some points. For example in a uniform density defined over $[0,1/10]$ the density will be $10$ between $0$ and $1/10.$

coffeemath
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