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If $X$ be an uniform distribution defined on $(-k,k)$, then the value of $k$ for so that : $$P(|X|<1) = P(|X|>2)$$

I began by defining the $p.d.f$ of the Uniform function namely: $$ f(x) = \frac{1}{2k}$$

Then we can re-write the LHS of the probability as : $$ P(|X|<1) = P(-1<X<1)$$

And the RHS : $$ P(|X|>2) = 1-P(|X|<2) \ = 1-P(-2<X<2) $$

Now solving the probability by $integration$ we get:

$$\int_{-1}^{1} \frac{1}{2k} dx \ = 1 - \int_{-2}^{2}\frac{1}{2k}dx$$

$$\frac{1}{k} = 1 - \frac{2}{k}$$ And therefore $$k=3$$

Is my method and answer correct?

Kalvin
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    To me it seems correct – Fiodor1234 Sep 15 '21 at 20:08
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    What you need is a method to check your work. Probability problems are often susceptible to quick, easy, informative verification by simulation. For instance, this `R` command prints estimates of the two probabilities based on a million draws of $X$: `(function(x) c(mean(abs(x) < 1), mean(abs(x) > 2))) (runif(1e6, -3, 3))`. It runs in a fraction of a second and will tell you whether you're even close to correct. – whuber Sep 15 '21 at 20:42
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    Another approach to checking your work (besides the highly useful approach of simulation) would be to draw a picture. However, you can also simplify things by working directly with $V=|X|$ rather than with $X$ itself while doing so. – Glen_b Sep 15 '21 at 23:56

1 Answers1

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Your result is correct, but the method has a small flaw. Your density should be $$ f(x) = \frac{1}{2k} \cdot I_{(-k,k)} (x) $$ This didn't change anything, since both your integration areas lie in $(-3,3)$, but in general this could fail.

Ben
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Jannis H.
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  • What is the $I_{-k,k}$ that you have included in the density? And how would it effect the method if the integration were to lie between say (0,k)? – Kalvin Sep 16 '21 at 03:01
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    Kalvin, please search our site for [indicator function](https://stats.stackexchange.com/search?q=indicator+function). – whuber Sep 16 '21 at 12:56
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    @Kalvin The function $I_{(-k,k)}(x)$ is the indicator function which is $1$ if $x \in (-k,k)$ and $0$ otherwise. – Jannis H. Sep 16 '21 at 15:28