Assume that we have a random variable $W = \max({X,Y})$ and that we would like to find the pdf of $W$. This is what I have done.

$$ F_W(w)= \mathbb{P}[ W\leq w]=\mathbb{P}[ \max({X,Y})\leq w]=\mathbb{P}[ X\leq w]\mathbb{P}[Y\leq w]= F_X(w)F_y(w) $$ then the pdf is $$f_W(w) = \frac{dF_W(w)}{dw}=\frac{d (F_X(w)F_y(w))}{dw}= f_x(w)F_y(w)+ f_y(w)F_x(w)$$

Is my reasoning correct?

What if one want to find the distribution of $W = \min({X,Y})$?

  • $\begingroup$ Assuming that $X$ and $Y$ are independent and both continuous random variables. Yes. $\endgroup$ Jan 22, 2015 at 3:03
  • $\begingroup$ thanks, i want to find the min of the two variables, can you guide me through the process? $\endgroup$
    – Henry
    Jan 22, 2015 at 3:12
  • $\begingroup$ Same principle, only you want $f_Z(z) = \frac{\mathrm d\;\;}{\mathrm d\, z}\Big( 1-\Bbb P(\min(X,Y)>z) \Big)$ $\endgroup$ Jan 22, 2015 at 3:23
  • $\begingroup$ Thanks for the answer, i dont know why i didnt think of it. $\endgroup$
    – Henry
    Jan 22, 2015 at 3:32

1 Answer 1


$\color{green}{\checkmark}\quad$ Your reasoning for the density function of $W=\max(X,Y)$ is correct. Apply the same reasoning to find that of the minimum, with minor modification.

Vis: Let $Z=\min(X,Y)$ such that $X$ and $Y$ are independent and both continuous random variables, with density functions: $f_X$ and $f_Y$. (And cumulative distributions $F_X, F_Y$).

$\begin{align} f_Z(z) & = \frac{\mathrm d\;}{\mathrm d z} \Bbb P(\min(X,Y)\leq z) \\[1ex] & = \frac{\mathrm d\;}{\mathrm d z} (1 - \Bbb P(\min(X,Y)\gt z)) \\[1ex] & = -\frac{\mathrm d\;}{\mathrm d z}\Big( \Bbb P(X>z)\,\Bbb P(Y>z) \Big) \\[1ex] & = -\frac{\mathrm d\;}{\mathrm d z} \Big(\big(1-F_X(z)\big)\big(1-F_Y(z)\big)\Big) \\[1ex] & = f_X(z)\Big(1-F_Y(z)\Big) + \Big(1-F_X(z)\Big)f_Y(z) \end{align}$

  • $\begingroup$ Hi @Graham could you give the intuition behind why subtract 1 - ℙ[min(𝑋,𝑌)>z] vs. just ℙ[max(𝑋,𝑌)≤𝑤] ? $\endgroup$
    – Googme
    Oct 17, 2020 at 6:36
  • $\begingroup$ The event of the minimum being less than a number is the event of at least one variable being so.$$\{\min(X,Y)< n\}=\{X<n\}\cup\{Y<n\}$$However, the event of the minimum being greater than a number is the event of both variables being greater than that number. $$\{\min(X,Y)> n\}=\{X>n\}\cap\{Y>n\}$$ The probability for the intersection of two independent events equals the product of the probabilities for each event. So working with this is useful. $\endgroup$ Oct 18, 2020 at 6:26
  • $\begingroup$ For maximum, this is the other way around. The event of the maximum being less than a number is the event of both variable being so. Thus this is the best to work with..$$\{\max(X,Y)< n\}=\{X<n\}\cap\{Y<n\}$$ $\endgroup$ Oct 18, 2020 at 6:30

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