The variable X has pdf $$f(x) = \frac18(6 - x)$$ for $$2 ≤ x ≤ 6$$

A sample of two values of X is taken. Denoting the lesser of the two values by Y, use the cdf of X to write down the cdf of Y. Hence obtain the pdf and mean of Y . Show that its median is approximately 2.64. (The median is the point m for which P(Y ≤ m) = 0.5.)

This is an exercise of Probability theory. It has been bothering me for a long time. Really thank you for helping me.

I am stuck in the first step:using the cdf of X to write down the cdf of Y.

I tried the Inverse function of cumulative distribution function, but it didn't seem to help.

  • $\begingroup$ At which point are you stuck? E.g. the first step would be to recognize that $P(Y \le a) = P(X_1 \le a \cap X_2 \le a) = P(X_1 \le a) \cdot P(X_2 \le a) = cdf_X(a)^2$... $\endgroup$ – antkam Apr 22 at 4:17
  • $\begingroup$ @LeeDavidChungLin Thanks, I am trying to learn it. $\endgroup$ – gong.y Apr 22 at 5:17
  • $\begingroup$ @gong.y - Obviously you are new here, so FYI the common etiquette is to not just list the problem but also tell people what you have tried, where you are stuck. This problem is a good example -- it has multiple steps and I had to guess where you are stuck. :) $\endgroup$ – antkam Apr 22 at 5:26
  • $\begingroup$ @antkam Yes, you are totally right. I get it now. $\endgroup$ – gong.y Apr 22 at 5:28
  • $\begingroup$ @antkam I use the square of the cdf of X as the cdf of Y for [2, 6]. But the total probability is not 1. I don't know why... $\endgroup$ – gong.y Apr 22 at 6:19

In order to follow the hint of antkam, and use the CDF of $Y$, it is necessary first to find the CDF of $X$. What we have now is the PDF of $X$. We must first integrate that with respect to $x$:

$$ F_X(x) = \int \frac18 (6-x) \, dx = \frac18 \left(6x-\frac12x^2\right) + C = \frac34 x - \frac{1}{16}x^2 + C, \qquad 2 \leq x \leq 6 $$

We need $F_X(2) = 0$, which yields $C = -\frac54$ and then

$$ F_X(x) = \frac34 x - \frac{1}{16}x^2 - \frac54 $$

As a check, we observe that $F_X(6) = \frac92 - \frac94 - \frac54 = 1$, as it should. You should now be able to identify the proper CDF for $Y$, and to determine the median for $Y$. Be aware that

$$ F_Y(y) = P(Y \leq y) = 1-P(Y > y) = 1-[P(X > y)]^2 = 1-[1-F_X(x)]^2 $$

so the hint of antkam is slightly off.

  • $\begingroup$ Thank you very much! $\endgroup$ – gong.y Apr 22 at 7:08

Hint: Let $X_1,X_2$ be the two samples, so $Y=\min\{X_1,X_2\}$. $$\mathsf P(Y\leqslant y)~{=1-\mathsf P(y<Y)\\[2ex]=~1-\mathsf P(y<\min\{X_1,X_2\})\\[2ex]=1-\mathsf P(y<X_1)~\mathsf P(y<X_2)}$$

  • $\begingroup$ Thank you very much! sorry I can only accept one answer. $\endgroup$ – gong.y Apr 22 at 7:07

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