4
$\begingroup$

Lets assume I have samples from 5 unique populations. Let's also assume I have a mean and standard deviation from each of these populations, they are normally distributed and completely independent of one another.

How can I estimate the probability that a sample of one of the populations will be greater than a sample from each of the other 4 populations?

For a example, if I have 5 types of fish (the populations) in my pond, such as bass, catfish, karp, perch and bluegill, and i'm measuring the lengths (the variables) of the fish, how do can I estimate the probability that the length of a bass I catch will be greater than the length of all the other types of fish? I think I understand how to compare 2 individual populations but can't seem to figure out how to estimate probability relative to all populations. As opposed to the probability of the bass to a catfish, and then a bass to a karp, etc., I'd like to know if its possible to reasonably estimate the probability of the length of the bass being greater that the lengths of all other populations.

Any help would be greatly appreciated! Thanks!

$\endgroup$

1 Answer 1

6
$\begingroup$

Edit: I believe my original solution is incorrect. I treated the events [koi > catfish] and [coy > karp] as independent when they are certainly not.

New answer

$$ \begin{aligned} P(Y>\max\{X_1,...,X_n\})&=P(Y>X_1,...,Y>X_n)\\ &=\int_{-\infty}^{\infty} P(Y>X_1,...,Y>X_n|Y=y) f_Y(y)dy\\ &=\int_{-\infty}^{\infty} \prod_{i=1}^n \left[ P(Y>X_i|Y=y) \right]f_Y(y)dy\\ &=\int_{-\infty}^{\infty} \prod_{i=1}^n \left[ \Phi \left( \tfrac{y-\bar{x}_n}{\sigma_{x_n}} \right) \right]f_Y(y)dy \end{aligned} $$

I do hope that someone can provide a better solution, as the above expression seems mismatched with the relative simplicity of the question.

The original (wrong!) answer

Let $Y$ represent the length of a fish from the population of interest, such as bass, and $X_i$ represent the length of fish from another population $i$, such as karp or catfish. You want to calculate the probability that the bass is longer than the longest non-bass fish. That is equivalent to the probability that the bass is longer than the carp, and the bass is longer than the catfish, and the bass is longer than the perch, etc. $$P(Y>\max\{X_1,...,X_n\})=P(Y>X_1,...,Y>X_n)$$

Because the lengths of your fish are independently distributed, the probability of all of these events happening is the product of the individual probabilities.

$$P(Y>X_1,...,Y>X_n) =\prod_{i=1}^{n} P(Y>X_i)$$

So the probability that bass is longer than all of your other fish is found by multiplying the probabilities that the bass is larger than each other type of fish.

That leaves only the problem of calculating the probability that a fish from one normal distribution is longer than a fish from another normal distribution. That is, $P(Y>X_i)$. To calculate this probability we rewrite it (ignoring the subscript) in the form $$P(Y>X)=P(Y-X>0)$$

Thankfully, the distribution of $Y-X$ is simple in the case where $X$ and $Y$ are normally distributed. That is, $X \sim N(\mu_{X},\sigma_{X})$ and $Y \sim N(\mu_{Y},\sigma_{Y})$. We can use the following facts:

  • Any linear combination of independent normal random variables (ie. $aX+bY$) is itself a normal random variable.
  • $\mathbb{V}(aX+bY)=a^2\mathbb{V}(X)+b^2\mathbb{V}(Y)$ for any uncorrelated random variables $X$ and $Y$.
  • $\mathbb{E}(aX+bY) = a\mathbb{E}(X)+b\mathbb{E}(Y)$ for any random variables $X$ and $Y$.

In this problem, the difference in the lengths of the two fish $D=Y-X=(1)X+(-1)Y$ is a linear combination of the two lengths, $X$ and $Y$. Therefore, using the facts above, we find that the distribution of the difference in lengths is

$$D\sim N(\mu_Y-\mu_X,\sigma^2_X+\sigma^2_Y)$$

The probability that this difference is greater than zero is

$$P(D>0)=1-P(D<0)=1-F_D(0)=1-\Phi \left(\frac{0-\mu_D}{\sigma_D} \right)$$

In terms of $X$ and $Y$ this is

$$P(Y-X>0)=1-\Phi \left(\frac{\mu_X-\mu_Y}{\sqrt{\sigma^2_X+\sigma^2_Y}}\right)$$

The final solution, in all its glory, would then be:

$$P(Y>\max\{X_1,...,X_n\})=\prod_{i=1}^{n} 1-\Phi \left(\frac{\mu_{X_i}-\mu_Y}{\sqrt{\sigma^2_{X_i}+\sigma^2_Y}}\right)$$

$\endgroup$
5
  • 1
    $\begingroup$ Presumably your operator "$\cap$" means ordinary multiplication of numbers, because both its arguments (being probabilities) are numbers. Maybe there's a typo there? "This extends to" hides the content of the answer--it needs elaboration. The meaning of "alternatively" is not evident and so needs elaboration, too. $\endgroup$
    – whuber
    Jul 2, 2020 at 20:05
  • 1
    $\begingroup$ Thanks @whuber. Hopefully, the edited answer is clearer. $\endgroup$
    – Ryan Volpi
    Jul 2, 2020 at 20:45
  • 1
    $\begingroup$ It is, thank you (+1). I can't help thinking, though, that the OP might welcome some words about how the individual probabilities $P(Y\gt X_i)$ might be estimated or calculated. $\endgroup$
    – whuber
    Jul 2, 2020 at 20:59
  • 1
    $\begingroup$ One thing i'm struggling to understand, is after I find the product that the bass is larger than the karp, the catfish, etc., I do the same for each fish (the karp being larger than all others, the catfish being larger than all others, etc.). Wouldn't the sum of the probabilities of each fish being larger than all others be equal to 1? i'm not getting anywhere close to that, maybe i'm not understanding why it wouldn't equal 1? Surely one of the fish will be larger than all others? I can provide numbers and show what i'm coming up with if that helps. $\endgroup$
    – mc_chief
    Jul 9, 2020 at 15:43
  • 1
    $\begingroup$ @mc_chief Thank you for that excellent observation. My answer is very likely mistaken. I believe I treat the case where [koi > catfish] and [coy > karp] are independent events. In reality, they are not. I'll correct this in a new answer ASAP. $\endgroup$
    – Ryan Volpi
    Jul 9, 2020 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.