0
$\begingroup$

Suppose I am drawing coloured balls from a bag.

The ball can be red, green or blue.

The probabilities of drawing a red, green or blue bag are uncertain, but I have confidence bounds for the probabilities.

I am 90% confident that the probability of drawing a red ball is 45%-55%. The tails are symmetric i.e. there is a 5% change that the probability is below 45%, and 5% chance that it is above 55%.

I am 90% confident that the probability of drawing a blue ball is 25%-35%. Again, the tails are symmetric.

From the above information, what can I infer about the probability of drawing a green ball?

I've had a read about the Dirichlet distribution, but it doesn't seem quite right for this. What else should I look at?

$\endgroup$

2 Answers 2

1
$\begingroup$

We can explore the Dirichlet family of distributions to check if it contains some distribution satisfying your conditions. So, we have:

$$P\sim\text{Dir}(\alpha_1,\alpha_2,\alpha_3)$$

It is reasonable to impose $\mathbb E(P)=(0.50, 0.30, 0.20)$, which implies gives us: $$(\alpha_1,\alpha_2,\alpha_3)=(0.5x,0.3x,0.2x)$$ For some positive number $x$. Now it only remains to find $x$. I have written the following R code to find what value of $x$ makes the central interval for $P_1$ have a length of $0.10$:

for (x in seq(1, 500, 0.001)){
  alpha = c(0.5,0.3,0.2)
  interval <- stats::qbeta(p = c(0.05, 0.95), shape1 = x*alpha[1], shape2 = x*(alpha[2]+alpha[3]))
  if(interval[2]-interval[1] < 0.10){
    print(x)
    print(interval)
    break
  }
}

This gives us $x\approx270$. If we write a similar code to assert the interval for $P_2$ has length $0.10$, we get a slightly different value of $x$:

for (x in seq(1, 500, 0.001)){
  alpha = c(0.5,0.3,0.2)
  interval <- stats::qbeta(p = c(0.05, 0.95), shape1 = x*alpha[1], shape2 = x*(alpha[2]+alpha[3]))
  if(interval[2]-interval[1] < 0.10){
    print(x)
    print(interval)
    break
  }
}

This gives us $x\approx226$. OK, so since the values of $x$ are different, the Dirichlet distribution does not allow your conditions to work exactly. But can we tweak it to make your conditions work with a good approximation? Let's try using $x=\frac{270+226}{2}=248$ and see what happens:

> x <- 248
> stats::qbeta(p = c(0.05, 0.95), shape1 = x*a[1], shape2 = x*(a[2]+a[3]))
[1] 0,4478658 0,5521342
> stats::qbeta(p = c(0.05, 0.95), shape1 = x*a[2], shape2 = x*(a[1]+a[3]))
[1] 0,2531545 0,3486831
> stats::qbeta(p = c(0.05, 0.95), shape1 = x*a[3], shape2 = x*(a[1]+a[2]))
[1] 0,1597154 0,2430401

OK, so $x=248$ gives us the distribution $\text{Dir}(124,74.4,49.6)$, which gives us $90\%$ intervals $(44.7\%-55.2\%)$ for $P_1$, $(25.3\%,34.9\%)$ for $P_2$ and $(16.0\%,24.3\%)$ for $P_3$.

TL;DR

  • The Dirichlet distribution is a good choice for the problem at hand
  • The adequate set of parameters is something close to $\alpha=(124,74.4,49.6)$
  • The probability of drawing a green ball is $\mathbb E[P_3]=0.20$ (this part follows from the other probabilities being $50\%$ and $20\%$, in average) and a $90\%$ interval around it is $16.0\%,24.3\%$ (AFAIK this part depends on using the Dirichlet distribution)
$\endgroup$
1
  • $\begingroup$ Nicely written and very well explained. Thank you @PedroSebe. $\endgroup$
    – user847663
    Commented Sep 22, 2020 at 15:11
1
$\begingroup$

If we call the probabilities as $P,Q,R$ for green, red and blue respectively, and assume the marginals given are correct, we can do the following w/o assuming anything about the joint PDF (i.e. Dirichlet or not):

$$P(\text{Green})=\int_{0}^1 P(\text{Green}|P=p)f_P(p)dp=\int_0^1 pf_P(p)dp=\mathbb E[P]$$

So, it is the expected value of $P$. If the marginal is symmetric around $50\%$, assuming the mean is defined, probability of drawing a green ball from the urn will be $0.5$.

$\endgroup$

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.