1. Indicator function

Hey,

The indicator function is defined as:

$\displaystyle \mathbb{1}_{A}(x)=1$ if $\displaystyle x \in A$
$\displaystyle \mathbb{1}_{A}(x)=0$ if $\displaystyle x \notin A$

I want to show that: $\displaystyle \mathbb{1}_{A\cup B}(x)= \min (\mathbb{1}_{A}(x) + \mathbb{1}_{B}(x),1)$.

Here what I have done:

For every $\displaystyle x$ we have that $\displaystyle x \in A \cup B$ means that $\displaystyle x \in A$ or $\displaystyle x \in B$.
This means that $\displaystyle \mathbb{1}_{A}=1$ or $\displaystyle \mathbb{1}_{B}=1$.
How do I argue from here? I know the addition rule for mutual exclusive events. Maybe I am overthinking this?

2. Re: Indicator function Originally Posted by detalosi Hey,

The indicator function is defined as:

$\displaystyle \mathbb{1}_{A}(x)=1$ if $\displaystyle x \in A$
$\displaystyle \mathbb{1}_{A}(x)=0$ if $\displaystyle x \notin A$

I want to show that: $\displaystyle \mathbb{1}_{A\cup B}(x)= \min (\mathbb{1}_{A}(x) + \mathbb{1}_{B}(x),1)$.

Here what I have done:

For every $\displaystyle x$ we have that $\displaystyle x \in A \cup B$ means that $\displaystyle x \in A$ or $\displaystyle x \in B$.
This means that $\displaystyle I_{A}=1$ or $\displaystyle \mathbb{1}_{B}=1$.
How do I argue from here? I know the addition rule for mutual exclusive events. Maybe I am overthinking this?
Mutually exclusive has nothing to do with it. Let's call $\min (I_{A}(x) + I_{B}(x),1)~f(x)$ to save writing. So what you have above with $x \in A\cup B$ is that $I_A(x) = 1$ or $I_B(x) = 1$. What does that tell you about the value of $f(x)$ for that $x$? Does that agree with $I_{A\cup B}(x)$? Now look at the same idea for $x \notin A\cup B$.

3. Re: Indicator function

If $\displaystyle x \in A \cup B$ then $\displaystyle \mathbb{1}_{A}=1$ or $\displaystyle \mathbb{1}_{B}=1$. Therefore $\displaystyle f(x)=\min(\mathbb{1}_{A}+\mathbb{1}_{B},1)=(1+1,1) =1$ (this matches with the definition of $\displaystyle \mathbb{1}_{A\cup B}$ when $\displaystyle x \in A \cup B$).

If $\displaystyle x \notin A \cup B$ then $\displaystyle \mathbb{1}_{A}=0$ or $\displaystyle \mathbb{1}_{B}=0$. Therefore $\displaystyle f(x)=\min(\mathbb{1}_{A}+\mathbb{1}_{B},1)=(0+0,1) =0$ (this matches with the definition of $\displaystyle \mathbb{1}_{A\cup B}$ when $\displaystyle x \notin A \cup B$).

The above method simply seems to be a verification-method and not a derivation in the sense that one starts with $\displaystyle \mathbb{1}_{A\cup B}$ and then arrives at $\displaystyle \min(\mathbb{1}_{A}+\mathbb{1}_{B},1)$. Is there another way to go about the above; a more direct proof maybe?

4. Re: Indicator function Originally Posted by detalosi If $\displaystyle x \in A \cup B$ then $\displaystyle \mathbb{1}_{A}=1$ or $\displaystyle \mathbb{1}_{B}=1$. Therefore $\displaystyle f(x)=\min(\mathbb{1}_{A}+\mathbb{1}_{B},1)=(1+1,1) =1$ (this matches with the definition of $\displaystyle \mathbb{1}_{A\cup B}$ when $\displaystyle x \in A \cup B$).

If $\displaystyle x \notin A \cup B$ then $\displaystyle \mathbb{1}_{A}=0$ or $\displaystyle \mathbb{1}_{B}=0$.
That isn't quite right. If you think a bit more about it you may see that you want "and" there, which you need to get the conclusion. If they weren't both $0$ then $f(x)$ wouldn't be $0$.

Therefore $\displaystyle f(x)=\min(\mathbb{1}_{A}+\mathbb{1}_{B},1)=(0+0,1) =0$ (this matches with the definition of $\displaystyle \mathbb{1}_{A\cup B}$ when $\displaystyle x \notin A \cup B$).

The above method simply seems to be a verification-method and not a derivation in the sense that one starts with $\displaystyle \mathbb{1}_{A\cup B}$ and then arrives at $\displaystyle \min(\mathbb{1}_{A}+\mathbb{1}_{B},1)$. Is there another way to go about the above; a more direct proof maybe?
Once it is corrected, it shows that the right and left sides are equal for all $x$. You can hardly get more direct than that.

5. Re: Indicator function

Yes, you are quite right about the correction.

Assume that you had no information about the right-hand-side. How would you arrive at the information on the right-hand-side?

6. Re: Indicator function Originally Posted by detalosi Yes, you are quite right about the correction.

Assume that you had no information about the right-hand-side. How would you arrive at the information on the right-hand-side?
I would probably notice that if $x\in A\cup B$ and I calculate $I_A(x)+I_B(x)$ I get $1$ unless $x$ is also in $A\cap B$ in which case I get $2$. So I would think about how to modify $I_A(x)+I_B(x)$ to make it work in that case.

7. Re: Indicator function

Let x be any point in the universal set. Then

case 1) $\displaystyle x\notin A\cup B$ so $\displaystyle i_{A\cup B}= 0$
Then x is not in A, so $\displaystyle i_A= 0$, and x is not in B, so $\displaystyle i_B= 0$. $\displaystyle i_A+ i_B= 0+ 0= 0$ so $\displaystyle min(i_A+ i_B, 1)= min(0, 1)= 0= i_{A\cup B}$.

case 2) $\displaystyle x\in A\cup b$ so $\displaystyle i_{A\cup B}= 1$
case 2a) $\displaystyle x\in A$ but $\displaystyle x\notin B$
Then x is in A, so $\displaystyle i_A= 1$, but x is not in B, so $\displaystyle i_B= 0$. $\displaystyle i_A+ i_B= 1+ 0= 1$ so $\displaystyle min(i_A+ i_B, 1)= min(1, 1)= 1= i_{A\cup B}$.

case 2b) $\displaystyle x\notin A$ but $\displaystyle x\in B$
Then x is in B, so $\displaystyle i_b= 1$, but x is not in A, so $\displaystyle i_A= 0$. $\displaystyle i_A+ i_B= 0+ 1= 1$ so $\displaystyle min(i_A+ i_B, 1)= min(1, 1)= 1= i_{A\cup B}$.

case 2c) $\displaystyle x\in A$ and $\displaystyle x\in B$
Then x is in A, so $\displaystyle x_A= 1$, and x is in B, so $\displaystyle i_B= 1$. $\displaystyle i_A+ i_B= 1+ 1= 2$ so $\displaystyle min(i_A+ i_B, 1)= min(2, 1)= 1= i_{A\cup B}$.

Notice that it is only this last case that requires "$\displaystyle min(i_A+ i_B, 1)$" rather than just "$\displaystyle i_A+ i_B$".

8. Re: Indicator function Originally Posted by HallsofIvy Let x be any point in the universal set. Then
case 2c) $\displaystyle x\in A$ and $\displaystyle x\in B$
Then x is in A, so $\displaystyle x_A= 1$, and x is in B, so $\displaystyle i_B= 1$. $\displaystyle i_A+ i_B= 1+ 1= 2$ so $\displaystyle min(i_A+ i_B, 1)= min(2, 1)= 1= i_{A\cup B}$.
Notice that it is only this last case that requires "$\displaystyle min(i_A+ i_B, 1)$" rather than just "$\displaystyle i_A+ i_B$".
We can also note that $x\in A~\&~x\in B$ is $I_{A\cap B}=I_{A}+I_{ B}-I_{A}\cdot I_{ B}$