# What Distribution Do I need?

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?

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)
• Nicely written and very well explained. Thank you @PedroSebe. Commented Sep 22, 2020 at 15:11

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 $$0.5$$.