Let $X_1, X_2, ...,X_n$ denote independent and identically distributed (iid) random variables (r.v) each with pdf $f_x(x)$. The (homework) problem asks me to consider the function:

Y = min{$X_1,X_2,...,X_n$}

And the goal is to find the PDF of Y.

Since the min{} function returns one of the random variables, isn't the PDF just $f_x(x)$? (He drops the hint that we should find the CDF of of Y and then take the derivative, which is rather involved, ergo I don't think I understand how min{} affects PDF. Can someone tell me more about how min{} may affect the r.v's?


Suppose that the random variables are iid uniform over the interval $[0,1]$. The pdf is a flat line because each of the $X_{i}$ are equally likely to be anywhere between $[0,1]$. However, if you collect $n$ of them and look at the minimum, it is more likely to, for example, be towards the lower half of the interval than the upper half so it's pdf should have more area towards $0$ and less up by $1$. i.e. It won't be a flat line anymore!

We can represent this problem using the following experiment:

Draw a number line and put $y$ on it. $y$ can be any value but thinking about the range of the function $min(X_{1},...,X_{n})$, $y$ should sensibly be a positive real number in the interval [0,1]. Now imagine the different number of ways you can drop $n$ data points around $y$ so that the minimum value of all points (when considered together) is below $y$. There are a lot of ways to do this but you must keep in mind that each point must be located between the interval [0,1] since the probability of them appearing elsewhere outside of the closed interval [0,1] is 0.

Your instructor is giving you a good hint. It is almost always easiest to use cdfs when trying to find the distribution of mins and maxes. Note that

$$ F_{Y(y)} = P(Y \leq y) = P(\min (X_{1}, X_{2}, \ldots, X_{n}) \leq y). $$

The idea is to relate this to probabilities involving the individual $X_{i}$ since you know things about them.

Now consider writing: $$ P(\min (X_{1}, X_{2}, \ldots, X_{n}) \leq y) = 1 - P(\min (X_{1}, X_{2}, \ldots, X_{n}) > y). $$

The only way to arrange your points so that $\min (X_{1}, X_{2}, \ldots, X_{n}) > y$ is to put them all above $y$.

So, $$ \begin{array}{lcl} F_{Y(y)} &=& P(Y \leq y) = P(\min (X_{1}, X_{2}, \ldots, X_{n}) \leq y)\\ &=& 1-P(\min (X_{1}, X_{2}, \ldots, X_{n}) > y)\\ &=& 1-P(X_{1}>y, X_{2}>y, \ldots, X_{n}>y) \end{array} $$

Now use the fact that the $X_{i}$ are independent to break that probability up into a product. Use the fact that they are identically distributed in order to write it as one probability to the $n$th power.

You are almost there since you can relate that one probability to the cdf of your original distribution!

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  • $\begingroup$ Hi, I don't understand how Y shouldn't have a uniform distribution. Can you devise an example of a random experiment Y (minding that Y = min{$X_1,X_2,...,X_n$} where you interpret what Y, $X_1,X_2,...,X_n$, correspond to? e.g, Let $X_i$ be the outcome of choosing a real number in the interval [0,1] such that each real number has equal likelihood of being chosen. $\endgroup$ – Minh Tran Oct 23 '15 at 16:00
  • $\begingroup$ Suppose that you have a random sample $X_{1},X_{2}$ of size $2$ from the discrete (just as an example) uniform distribution on $\{0,1\}$. Let $Y = \min (X_{1},X_{2})$. Then $Y$ takes on values in $\{0,1\}$ but not in an equally likely way. In particular, $P(Y=1) = P(\min(X_{1},X_{2})=1)=P(X_{1}=1,X_{2}=1) \stackrel{indep}{=} (1/2) \cdot (1/2) = 1/4 \ne 1/2$. $\endgroup$ – Rubarb Oct 24 '15 at 1:56
  • $\begingroup$ @MinhTran I appreciate that you liked this answer, but I don't understand why you did such violence to it in editing it. The original answer, after the first paragraph, applied to any distribution with a pdf; the edited version takes a part of the answer that may or may not have been necessary, removes it from the context where it at least had some point, makes it about a special case (uniform distribution over [0,1]), and then leaves it hanging without concluding anything from it. If I treated your writing as cavalierly as you treated Rubarb's, I would just roll back the edit. $\endgroup$ – David K Jan 10 '17 at 4:01
  • $\begingroup$ @DavidK Rereading my question, I agree with you that a good example shouldn't assume a particular pdf, as Rubarb had originally mentioned. What helped me understand his response was reading the first and third paragraph in succession, skipping the paragraph in between that was less exposition, and more about how to solve the problem. This is why I suggested moving the second paragraph closer to the bottom. $\endgroup$ – Minh Tran Jan 10 '17 at 20:02
  • $\begingroup$ I also agree with you that the edit to the (now) second paragraph did remove it out of the context of the original question. I will re-edit this. $\endgroup$ – Minh Tran Jan 10 '17 at 20:06

Take for example the $X_i$ uniform on $[0,1]$, and $n=100$. The minimum $Y$ of the $X_i$ is likely to be quite small, and will clearly have mean quite a bit smaller than $\frac{1}{2}$. Its distribution will not be the same as the distribution of the $X_i$.

Let the density function of the $X_i$ be $f_X(x)$, and let $F_X(x)$ be their cdf. We will find the cdf $F_Y(y)$ of their minimum $Y$. Note $Y\gt y$ if and only if all the $X_i$ are greater than $y$. The probability that all the $X_i$ are greater than $y$ is $(1-F_X(y))^n$. Thus $$F_Y(y)=\Pr(Y\le y)=1-(1-F_X(y))^n.$$ If the $X_i$ have continuous distribution, then differentiating we get $$f_Y(y)=nf_X(y)(1-F_X(y))^{n-1}.$$

Remark: The distribution of the max $W$ of the $X_i$ is obtained in a similar but easier way. For $W\le w$ if and only if all the $X_i$ are $\le w$. This has probability $(F_X(w))^n$. Now, if the $X_i$ have continuous distribution, we can differentiate to find $f_W(w)=nf_X(w)(F_X(w))^{n-1}$.

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