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Say we have a random variable $X$ with some probability distribution $D$. For example, $D$ could be the normal distribution: $$\text{pdf}_X(x)=D(x)=\frac{1}{\sqrt {2\pi \sigma ^2}}\int_{-\infty}^{\infty}e^{1\frac{x^2}{2\sigma^2}}dx$$

Now let's say we have an arbitrary function $f(x)$. Then we can define a new random variable $Y=f(X)$.

Now my question is, what is the probability density function of $Y$? What is $\text{pdf}_Y$, given that we know $f(x),$ and $\text{pdf}_X$?

Or more generally: How do we even approach solving this problem? I don't even know where to start.

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1 Answer 1

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This is a question of transformation of random variable. If $X$ has a PDF $D(x)$ and $y=f(x)$ is a one to one function, the PDF of $y$ ($D'(Y)$) can be written as

\begin{equation} D'(y)=D(g^{-1}(y))\times\left| \frac{dx}{dy}\right|. \end{equation}

For more details go through the following text: http://math.arizona.edu/~jwatkins/f-transform.pdf.

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