Let $X$ be a continuous random-variable with probability distribution $f_X(x)$.

Let $Y=g(X)$, where $g(\cdot)$ is some transformation and we also know $f_Y(y)$.

Question How the probability distribution be affected if transformation is translated. i-e

If $Z=g(X-c)$ and we already know $f_Y(y)$ then how $f_Z(z)$ be defined in terms of $f_Y(y)$?

  • $\begingroup$ Is this for some subject? $\endgroup$ – Glen_b Nov 18 '14 at 5:12
  • $\begingroup$ No. I was trying to find the pdf of transformation Sin$(\theta)$ when $(\theta)$ itself is uniformly distributed. It seems that inversion of sin$(\theta)$ is not very straight forward as that of cos$(\theta)$, and that for Cos$(\theta)$ I do know the respective PDF. So, if I be able to know the effect of transformation then things become easy, as Sin$(\theta)=$Cos$(\theta- \pi/2)$ $\endgroup$ – kaka Nov 18 '14 at 5:19
  • $\begingroup$ Uniform on what range? $\endgroup$ – Glen_b Nov 18 '14 at 5:45
  • $\begingroup$ $\theta \sim \mathcal{U}(0,2 \pi)$ $\endgroup$ – kaka Nov 18 '14 at 5:50
  • $\begingroup$ It seems to be pretty straightforward to solve the original problem to me. But the two problems are so closely related you've already solved the problem. $\endgroup$ – Glen_b Nov 18 '14 at 5:50

For simplicity, suppose $g$ is one-to-one. So $X = g^{-1}(Y)$ and $X - c= g^{-1}(Z)$, and $g^{-1}(Y) = g^{-1}(Z) + c, $ and $Y = g(g^{-1}(Z) + c)$

$f_Y(y) = \frac{d}{dy} F_x(g^{-1}(y)) = f_X(g^{-1}(y))*(\frac{d}{dy}(g^{-1}(y))) = f_X(g^{-1}(y))*\frac{dx}{dy} $


$f_Z(z) = \frac{d}{dz} F_x(g^{-1}(z) +c) = f_X(g^{-1}(y))*(\frac{d}{dz}(g^{-1}(z)+c)) = f_X(g^{-1}(y))*\frac{dx}{dz} $

Thus, $f_Z(z) = f_Y(y)* \frac{dy}{dz} = f_Y(g(g^{-1}(z) + c)) * \frac{dg(g^{-1}(z) + c)}{dz}$

I am not sure this derivation is entirely correct, and welcome any commenter to add comments on errors if any.

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