Reviewing Basic Probability: Solving for $E(min(A,B))$

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In summary, the Expected Value is a measure of central tendency that represents the average outcome of a random variable. When calculating the Expected Value of the minimum of two random variables, we take the minimum of the expected values of each variable. This is useful in decision-making and risk analysis as it allows us to determine the most likely minimum outcome. However, the Expected Value of the minimum can be negative if one or both variables have negative outcomes. It has various real-world applications such as in insurance and finance, helping to calculate expected payouts and minimum returns.
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Jason4
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I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on $[0,1]$. What is $E(min(A,B))$ ?

$\displaystyle\int_{0}^{1}\int_{0}^{a}b\,db\,da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\,db\,da$

$=\displaystyle\int_{0}^{1}\frac{a^2}{2}\,da+\int_{0}^{1}a-a^2\,da$

$=1/6+3/6-2/6=1/3$
 
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  • #2
Jason said:
I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on $[0,1]$. What is $E(min(A,B))$ ?

$\displaystyle\int_{0}^{1}\int_{0}^{a}b\;db\;da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\;db\;da$

$=\displaystyle\int_{0}^{1}\frac{a^2}{2}\;da+\int_{0}^{1}a-a^2\;da$

$=1/6+3/6-2/6=1/3$

That is correct, though you could use a line of explanation at the begining:

\(E(\min(A,B))=E(A|A<B)+E(B|A\ge B)\)

CB
 
  • #3
Thanks Captain. I also need to find $E((A+B)^2)$ and $E(|A-B|)$.

For the first one:
\[E(A^2)+2E(A)E(B)+E(B^2)=1/3+2(1/4)+1/3)=7/6\]

Could you give me a hint for the second one?
 
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  • #4
$ \displaystyle \int_{0}^{1} \int_{0}^{1} |a-b| \ da \ db = \int_{0}^{1} \int_{b}^{1} (a-b) \ da \ db + \int_{0}^{1} \int_{0}^{b} -(a-b) \ da \ db =\frac{1}{3}$So $\frac{1}{3}$ is the expected distance between the two points. This sort of makes since since the expected value of min{A,B} is $\frac{1}{3}$, and the expected value of max{A,B} is $\frac{2}{3} $.
 
  • #5
So I could just calculate $E(max(A,B))-E(min(A,B))$ ?
 
  • #6
Jason said:
So I could just calculate $E(max(A,B))-E(min(A,B))$ ?
$\displaystyle \int_{0}^{1} \int_{0}^{1} |a-b| \ da \ db = \int_{0}^{1} \int_{b}^{1} (a-b) \ da \ db + \int_{0}^{1} \int_{0}^{b} -(a-b) \ da \ db$

$ \displaystyle = \Bigg( \int_{0}^{1} \int_{b}^{1} a \ da \ db + \int_{0}^{1} \int_{0}^{b} b \ da \ db \Bigg) - \Bigg( \int_{0}^{1} \int_{b}^{1} b \ da \ db + \int_{0}^{1} \int_{0}^{b} a \ da \ db \Bigg) $

$ \displaystyle = E(\max\{A,B\}) - E(\min\{A,B\})$
 
  • #7
Jason said:
I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on $[0,1]$. What is $E(min(A,B))$ ?

$\displaystyle\int_{0}^{1}\int_{0}^{a}b\,db\,da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\,db\,da$

$=\displaystyle\int_{0}^{1}\frac{a^2}{2}\,da+\int_{0}^{1}a-a^2\,da$

$=1/6+3/6-2/6=1/3$
It is correct. You might want to research 'Order Statistics'.
 

1. What is the definition of Expected Value?

The Expected Value is a measure of central tendency that represents the average outcome of a random variable. It is calculated by multiplying each possible outcome by its probability and summing the results.

2. How is the Expected Value of the minimum of two random variables calculated?

The Expected Value of the minimum of two random variables (A and B) is calculated by taking the minimum of the expected values of A and B. This means that the expected value of the minimum of A and B is equal to the smaller of the two expected values.

3. What is the significance of calculating the Expected Value of the minimum of two random variables?

Calculating the Expected Value of the minimum of two random variables is useful in scenarios where we are interested in finding the minimum of two outcomes, such as in decision-making or risk analysis. It allows us to determine the most likely minimum outcome.

4. Can the Expected Value of the minimum of two random variables be negative?

Yes, the Expected Value of the minimum of two random variables can be negative if one or both of the random variables have negative outcomes. It is important to consider the context of the problem when interpreting the result.

5. How can the Expected Value of the minimum of two random variables be used in real-world applications?

The Expected Value of the minimum of two random variables can be used in various real-world applications, such as insurance and finance. For example, in insurance, it can help calculate the expected payout in case of multiple claims, and in finance, it can help determine the expected minimum return on a portfolio of investments.

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