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For $0 \le a\le 1$ and $0\le b\le 1$ find the range of $a+b-ab$.

Here is what I've tried,

I noted that $0\le a+b\le 2$ and $0\le ab\le 1 \Rightarrow -1\le -ab\le 0$. And adding the inequalities gives, $-1\le a+b-ab\le 2$ Which doesn't look right.

But if I subtract $0\le ab\le 1$ from $0\le a+b \le 2$ this gives $0\le a+b-ab\le 1$. I'm wondering if the latter approach is correct and why the first one is wrong? I'm also interested in other approaches.

Etemon
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3 Answers3

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If $A \le B$ and $C \le D$ then $A+C \le B+D$. In that sense, inequalities can be “added.”

In your example, adding the two inequalities $0\le a+b\le 2$ and $-1\le -ab\le 0$ gives $-1\le a+b-ab\le 2$, which is correct. However, the bounds are not best possible because equality can not hold simultaneously in those inequalites.

But $A \le B$ and $C \le D$ does not imply $A-C \le B-D$, so inequalities can not be “subtracted” from each other in the same way as they are added. (Example: Subtract $1 \le 4$ from $3 \le 4$.)

In your example, subtracting $0\le ab\le 1$ from $0\le a+b\le 2$ happens to produce the correct $0\le a+b-ab\le 1$, but that is pure conincidence.

Here are some approaches for a correct solution:

  1. As already noted in the comments $$ a+b-ab=1-(1-a)(1-b) $$ and the product on the right is in the range $[0, 1]$, so that $0 \le a+b-ab \le 1$. Choosing $a=b=0$ and $a=b=1$ shows that these bounds are sharp.

  2. One can argue that the function $f(a, b) = a+b-ab$ is linear in each argument, so that both the minimum and the maximum on the square $[0, 1]^2$ are necessarily attained at one of the four vertices.

  3. Minimize/maximize the expression with respect to $b$ for fixed $a$, and then minimize/maximize it with respect to $a$: $$ a+b-ab = a + (1-a) \cdot b \ge a +(1-a) \cdot 0 = a \ge 0 $$ with equality for $a=b=0$, and $$ a+b-ab = a + (1-a) \cdot b \le a +(1-a) \cdot 1 = 1 $$ with equality for $b=1$.

Martin R
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I'm also interested in other approaches.

Consider two probability events A and B such that $a = P(A)$ and $b=P(B)$.

Then $a+b-ab = P(A \cup B)$.

Therefore if the range of $a$ and $b$ is $[0, 1]$ then the range of $a+b-ab$ is also $[0, 1]$.

Note that the previous sentence hides a two-way inclusion. Allow me to make it more explicit:

  • Let $0 \leq a \leq 1$ and $0 \leq b \leq 1$. Then I can find two random events $A$ and $B$ such that $P(A) = a$ and $P(B) = b$. Then $P(A \cup B) = a + b - ab$, which proves that $0 \leq a+b-ab \leq 1$. This proves that the range of $a+b-ab$ is included in $[0, 1]$.

  • Let $0 \leq y \leq 1$. Then I can find two random events $A$ and $B$ such that $P(A \cup B) = y$. Define $a = P(A)$ and $b = P(B)$; then $0 \leq a \leq 1$ and $0 \leq b \leq 1$ and $y = a+b-ab$. This proves that $[0, 1]$ is included in the range of $a+b-ab$.

Stef
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Inspired by @Martin R's answer, here is another approach:

Suppose $f(a,b) = a+b - ab$. Since both $a,b$ and $1-a, 1-b$ lies on $[0,1]$, We have $$f(1-a, 1-b) = 2-a-b-(1-b-a+ab) = 1-ab$$

Hence the range of the function is $[0,1]$.

Etemon
  • 6,437