I am trying to find analytically the distribution of $\sin(x)$, If x belongs to an exponential distribution,

  • 1
    $\begingroup$ Are you asking about the probability density function $f_Y(y)$ of the random variable $Y=\sin(X)$ where $X$ is another exponentially distributed random variable with density $f_X(x)=\lambda e^{-\lambda x}$, $x > 0$? If so, please edit your question accordingly. $\endgroup$ Commented Jul 17, 2019 at 16:52
  • $\begingroup$ The question is still unclear, because the notation "$f(x;\lambda)$" conventionally means $\lambda$ is a parameter of a family of distribution functions. The meanings of the equations "$\mu=\sigma$" and "$\lambda=1/\mu$" are obscure. Is your question trying to ask how $\sin(\exp(X))$ is distributed when $X$ has an exponential distribution with rate parameter $\lambda$? If so, then the thread at stats.stackexchange.com/questions/138763 will give you a good start on obtaining the answer. $\endgroup$
    – whuber
    Commented Jul 17, 2019 at 16:59

1 Answer 1


The cumulative distribution function (cdf) of a variable $X$ with an exponential distribution can be written

$$F_\lambda(x) = \Pr(X\le x) = 1 - e^{-\lambda x}.$$

Consequently, for any interval determined by $0\le a\le b,$ the chance $X$ lies in this interval is $$\Pr(a\lt X\le b) = F_\lambda(b)-F_\lambda(a) = e^{-\lambda a} - e^{-\lambda b}.$$

Let $T=\sin(X).$ Computing the CDF of $T$ using the graphical approach recommended at PDF of function of X, we find for $-1\lt t \le 1$ and $x = \arcsin(t)$ (a number between $-\pi/2$ and $\pi/2$) that the event $T\le t$ is the union of disjoint events

$$\mathcal{E}_i:2\pi j + \pi-x \le X \le 2\pi j + 2\pi + x,\ j=1,2,3,\ldots$$

together with the event

$$\mathcal{E}_0: 0 \le X \le x.$$

(This event is empty when $t\lt 0.$) Consequently the desired probability is the sum of probabilities of these events, which simplifies because these probabilities form a geometric sequence:

$$\eqalign{\Pr(T\le t) &= \Pr(X\in\mathcal{E}_0) + \sum_{j=1}^\infty\Pr(X\in\mathcal{E}_j) \\ &= 1-e^{-\lambda x} + \sum_{j=1}^\infty \left(e^{-\lambda(2\pi j + \pi -x)} - e^{-\lambda(2\pi j + 2\pi +x}\right) \\ &= 1-e^{-\lambda x} + \left(e^{-\lambda(\pi-x)} - e^{-\lambda(2\pi+x)}\right)\sum_{j=1}^\infty \left(e^{-\lambda 2\pi}\right)^j \\ &= 1-e^{-\lambda x} + \frac{e^{-\lambda(\pi-x)} - e^{-\lambda(2\pi+x)}}{1-e^{-\lambda 2\pi}} \\ &= 1-e^{-\lambda \arcsin(t)} + \frac{e^{-\lambda(\pi-\arcsin(t))} - e^{-\lambda(2\pi+\arcsin(t))}}{1-e^{-2\pi\lambda }}, }$$

with the first two terms $1-e^{-\lambda \arcsin(t)}$ not appearing when $\arcsin(t)\lt 0.$

Here are illustrations of the CDF, superimposed on the empirical distributions of samples of size 10,000. The theoretical curves (in red) coincide with the empirical distribution functions everywhere, demonstrating the correctness of these results. Beneath each CDF is a plot of the PDF of the underlying exponential distribution.

enter image description here


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