How to sample two values from a random variable X with the lesser to be a random variable Y?

Question

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.

Answer 1

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.

Answer 2

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