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Suppose that a random variable $X$ has a uniform distribution over the interval (0,1) i.e. $X$~U(0,1). Let $Y$ be a random variable such that $Y$ = |$X$-1/3|. What is the probability density function of $Y$?

How can this be solved?

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  • $\begingroup$ have you tried to just plot the new distribution function? $\endgroup$
    – user190080
    Commented Oct 29, 2015 at 18:11

1 Answer 1

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Just observe that - for example by plotting the distribution function of $Y$ - that we have first $$ Y\in(0,\frac23) \text{ simply because }X\sim U(0,1),X\in(0,1) $$ and therefore $$ F_Y(y)=\begin{cases}0,y\leq0 \\2y,0< y\leq\frac13 \\ y+\frac13,\frac13\leq y<\frac23 \\ 1,y\geq\frac23 \end{cases} $$ whch gives us the density function by a piecewise derivative as well $$ f_Y(y)=\begin{cases}0,y\leq0 \\2,0< y\leq\frac13 \\ 1,\frac13\leq y<\frac23 \\ 0,y\geq\frac23 \end{cases} $$ So the steps are really

  • how does $Y$ looks like, plot

  • use the properties of $X$ distribution to find the cdf of $Y$

  • simply compute the derivative (piecewise) and glue together
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