If X is uniformly distributed over $(-1, 1)$, find $P(|X| \gt \frac12 )$

I know what the pdf looks like, it is $0$ when $x \gt 1$ and $x \lt -1$, and when $-1 \lt x \lt 1$ it is a constant $\frac12$.
if the question was asking for $P(X \gt \frac12)$ I would do $\int_\frac12^1 \frac12 dx$ and get the answer $\frac14$, but because there is an absolute value around the X in the question I don't know what to do?
Do I have to calculate two integrals and add them?
$\int_\frac12^1 \frac12 dx$ and $\int_{-1}^{-\frac12} \frac12 dx$?

  • $\begingroup$ Yes, you should go with the approach you have mentioned. But clean up your integrals - they don't look right. $\endgroup$ – soakley Nov 13 '15 at 2:32
  • $\begingroup$ why $\int x dx$? lets start from here $\endgroup$ – Seyhmus Güngören Nov 13 '15 at 2:32
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    $\begingroup$ You are overcomplicating this. For uniform distribution, evaluating the probability $P(X \gt \frac{1}{2})$ only needs the integral $\int_{\frac{1}{2}}^1 \frac{1}{2} \;dx$, and instead of "integrating" per se, you only need the length of the interval $(1/2,1)$ in proportion to $(-1,1)$. Now consider the union of intervals where $|X|\gt \frac{1}{2}$. $\endgroup$ – hardmath Nov 13 '15 at 2:33
  • $\begingroup$ I fixed the intervals, I put the x there beause I got confused with the formula of expectation of a pdf $\endgroup$ – idknuttin Nov 13 '15 at 2:42
  • $\begingroup$ hardmath, the union of intervals where $|X| \gt \frac12$ is the length of the interval $(\frac12, 1)$ and the length of the interval $(-1, {-\frac12})$, this is the same thing as what I am saying in my integrals correct? $\endgroup$ – idknuttin Nov 13 '15 at 2:49

Yes.   You must integrate over the two disjoined intervals of the support where the absolute value of the random variable are greater than an half.   Linearity of Integration means you may sum the integrals over each.

$$\begin{align} \mathsf P(\lvert X\rvert > \tfrac 1 2) & = \displaystyle\int_{\{x\in(-1;1):\lvert x\rvert>1/2\}} f_X(x)\operatorname d x \\[1ex] & = \int_{(-1;-1/2)\cup(1/2;1)}\tfrac 1 2\operatorname d x \\[1ex] & = \int_{-1}^{-1/2} \tfrac 1 2\operatorname d x+ \int_{1/2}^1 \tfrac 12\operatorname d x \end{align}$$

Which, as hardmath commented, doesn't actually need any calculus to find.   This is because the constant density means we just need to compare interval lengths.   [Important!: the distribution must be uniform for this to work.]

$$\begin{align} \mathsf P(\lvert X\rvert > \tfrac 1 2) & = \frac{\lvert(-1;-1/2)\cap(1/2;1)\rvert}{\lvert(-1;-1/2)\cap(1/2;1)\rvert}{\lvert(-1;1)\rvert} \\[1ex] & = \tfrac 1 2 \end{align}$$

But there's nothing wrong with ensuring you've mastered the concept before moving onto non-uniform distributions.   Good work.

  • $\begingroup$ if I wanted to find the density function of random variable |X| would it simply be this? f(|X|) = \begin{cases} \frac12, & \text{if $x$ is in [0, 1)} \\ 0, & \text{else} \end{cases} $\endgroup$ – idknuttin Nov 13 '15 at 17:43
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    $\begingroup$ Not quite, @idknuttin Remember that the absolute signage folds two intervals into one. $$f_{\lvert X\rvert}(x) = \big(f_X(x)+f_X(-x)\big) \mathbf 1_{x\in [0;1)} = \begin{cases}1 & : x\in [0;1) \\ 0 &: \textsf{otherwise} \end{cases}$$ As a reality check: the integral of the pdf over the support must equal unity. $\endgroup$ – Graham Kemp Nov 13 '15 at 22:04

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