How to understand that $\min(a,b) < c \Leftrightarrow a < c \text{ or } b < c$ and $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$??

Question

In many probability related questions, I need to find the the probability of

$P[\min(X_1,X_2) < c]$

where $X_1, X_2$ are random variables

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However, I can never remember this basic relationship:

given any real numbers $a,b,c$, we have that $\min(a,b) < c \Leftrightarrow a < c \text{ or } b < c$.

Ditto, $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$.

I have looked up the explanation and people basically said this is just "logic".

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I guess my problem is not understanding what $\min(a,b) < c$ means.

To me, it means pick a number less than $c$ that is either $a$ or $b$, and show me the smaller of the two.

Aha, that must be where the "or" come from.

But wait a minute, it is also true that $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$.

In this case, we also have: pick a number less than $c$ that is either $a$ or $b$, and show me the larger of the two.

Now where would the "and" come from?

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All in all, is there a more rigorous way of showing that this is true?

Answer 1

I guess my problem is not understanding what $\min(a,b) < c$ means.

To me, it means pick a number less than $c$ that is either $a$ or $b$, and show me the smaller of the two.

Aha, that must be where the "or" come from.

Indeed, $\min\{a,b\}< c$ means that the lesser of the two values is less than $c$; so one of them is but the other might not be.   Therefore $a<c$ or $b<c$.

$$a\leq b<c,\\ a<c\leq b,\\ b<a<c,\\ b<c\leq a.$$

But wait a minute, it is also true that $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$.

In this case, we also have: pick a number less than $c$ that is either $a$ or $b$, and show me the larger of the two.

Now where would the "and" come from?

$\max\{a,b\}<c$ means that the greater of the two values is less than $c$, and if the greater is then the lesser also must be too.   Hence, it means that both values are less than $c$   Therefore $a<c$ and $b<c$.

$$a\leq b<c,\\b<a<c.$$

Answer 2

I think of $\min(a,b)<c$ to mean that "the smaller number of $a$ and $b$ is less than $c$. I think of $\max(a,b)<c$ to mean that "the largest number out of $a$ and $b$ is less than $c$. If you think of it this way, then it becomes quite clear: if the smallest number out of $a$ and $b$ is less than $c$ then this means at least one of them is less than $c$, so either $a<c$ or $b<c$. If the largest number out of $a$ and $b$ is less than $c$, then since it is the largest of the two, the other one is less than this maximum number, so they musts both be less that $c$.

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More rigorously:

Suppose $a<b$. Then $\min(a,b)=a$, and since $\min(a,b)<c$, we have that $a<c$.
Alternatively, $b<a$. Then $\min(a,b)=b$, and since $\min(a,b)<c$, we have that $b<c$.
(Also, if $a=b$, then $\min(a,b)=a=b<c$, so this still holds)
So we have determined that if $\min(a,b)<c$, then either $a<c$ or $b<c$.

Using similar reasoning you can determine that the backward implication also holds.

For the $\max$ case, the proof is very similar, but the word I have in bold above, alternatively, becomes also.

Answer 3

Visualising numbers on an axis, it's obvious:

• The first assertion means the leftmost of $a$ and $b$ is left of $c$ if and only if one of $a$ and $b$ is left of $c$.
• The second assertion means the rightmost of $a$ and $b$ is left of $c$ if and only if both $a$ and $b$ are left of $c$.