1
$\begingroup$

Let $X_1$, $X_2$, $X_3$ be independent, identically distributed continuous random variables from Exponential Distribution.

I want to find the probability that the median is closer to the smallest value $X_($$_1$$_)$ than it is to the largest value $X_($$_3$$_)$.

So far I have:

$F_2$($X_2$) = P($X_($$_2$$_)$ ≤ $X_2$) = ∫ 3! f(x) (F(x)$^($$^($$^n$$^-$$^1$$^)$$^/$$^2$$^)$/((n-1)/2)!) ((1-F(x))$^($$^($$^n$$^-$$^1$$^)$$^/$$^2$$^)$)/((n-1)/2)!)

Note: F(x) is the cdf, and f(x) is the pdf of Exponential Distribution

From this I will integrate with respect to $X_2$ with n = 3, to get the median. However, how do show I show that this median is closer to $X_($$_1$$_)$ than $X_($$_3$$_)$ ?

$\endgroup$
2
$\begingroup$

The event of interest is $$|X_{(2)} - X_{(1)}| < |X_{(2)} - X_{(3)}|.$$ Since $X_{(1)} \le X_{(2)} \le X_{(3)}$ by definition, this reduces to $$X_{(2)} - X_{(1)} < X_{(3)} - X_{(2)},$$ or $$X_{(2)} < \frac{X_{(1)} + X_{(3)}}{2}.$$ Therefore, $$\Pr[X_{(2)} < \tfrac{1}{2}(X_{(1)} + X_{(3)})] = \int_{x_1 = 0}^\infty \int_{x_3 = x_1}^\infty \int_{x_2 = x_1}^{(x_1+x_3)/2} f_{X_{(1)}, X_{(2)}, X_{(3)}}(x_1, x_2, x_3) \, dx_2 \, dx_3 \, dx_1,$$ where $$f_{X_{(1)}, X_{(2)}, X_{(3)}}(x_1, x_2, x_3) = 3! f_{X_1}(x_1) f_{X_2}(x_2) f_{X_3}(x_3) = 6e^{-(x_1+x_2+x_3)}$$ is the joint density of the order statistics. This is easy to evaluate: $$\begin{align*} \Pr[X_{(2)} < \tfrac{1}{2}(X_{(1)} + X_{(3)})] &= 6 \int_{x_1=0}^\infty \int_{x_3 = x_1}^\infty e^{-(x_1+x_3)} \left(e^{-x_1} - e^{-(x_1+x_3)/2}\right) \, dx_3 \, dx_1 \\ &= 6 \int_{x_1 = 0}^\infty \frac{1}{3}e^{-3x_1} \, dx_1 \\ &= \frac{2}{3}. \end{align*}$$ Here we have assumed without loss of generality that the samples are IID from an exponential distribution with mean $1$. I leave it as an exercise to show that the calculation does not depend on the choice of parameter.

$\endgroup$
  • $\begingroup$ Thanks a lot, this is clear to understand! Can I apply this to any other distribution? (assuming I sub in the appropriate marginal pdf's) $\endgroup$ – queence Feb 1 '18 at 20:09
2
$\begingroup$

I see that an answer is already posted above, but I thought it would be nice to give an answer that avoids computation altogether.

Think of it in terms of arrival times and use the memoryless property of the exponential. Note that once we get the smallest value $X_{(1)}$, the distribution of $X_{(2)} - X_{(1)}$ is the minimum of two i.i.d. exponentials, and therefore has rate $2\lambda$. Similarly, once we know $X_{(2)}$, $X_{(3)} - X_{(1)}$ is just exponential with rate $\lambda$.

Thus, the probability that the median is closer to $X_{(1)}$ than $X_{(3)}$ is the same as the probability that an exponential with rate $2\lambda$ is smaller than another exponential with rate $\lambda$. This probability is $2\lambda/(\lambda + 2\lambda) = 2/3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.