SOLUTION: A coin has a probability 𝑝 of landing heads. It is flipped 10 times, resulting in 8 heads. Using Bayesian inference with the following prior information
(a) No prior knowledge
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Question 1209943: A coin has a probability 𝑝 of landing heads. It is flipped 10 times, resulting in 8 heads. Using Bayesian inference with the following prior information
(a) No prior knowledge about 𝑝.
(b) The coin is likely fair (𝑝 around 0.5).
(c) The coin is likely biased (𝑝 around 0 or 1).
(d) 𝑝 can only take one of three values: 0.2, 0.7, or 0.9.
(e) 𝑝 is restricted to the range 0.4 ≤ 𝑝 ≤ 0.9.
, answer the following questions:
(1) What is the estimated value of 𝑝?
(2) Is 𝑝 greater than 0.6?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem using Bayesian inference for each prior.
**1. Likelihood Function**
The likelihood function is the probability of observing 8 heads in 10 flips given a probability $p$ of heads. This follows a binomial distribution:
$L(p) = \binom{10}{8} p^8 (1-p)^2 = 45 p^8 (1-p)^2$
**2. Bayesian Inference**
Bayesian inference combines the likelihood function with a prior distribution to obtain a posterior distribution.
**a) No Prior Knowledge (Uniform Prior)**
* Prior: $P(p) = 1$ for $0 \le p \le 1$ (uniform distribution)
* Posterior: $P(p | \text{data}) \propto L(p) P(p) = 45 p^8 (1-p)^2$
* To find the estimated value of $p$, we need to find the maximum of the posterior. This is equivalent to maximizing the likelihood since the prior is uniform.
* To find the maximum, we take the derivative of $L(p)$ and set it to 0:
* $L'(p) = 45 (8 p^7 (1-p)^2 - 2 p^8 (1-p)) = 0$
* $8 (1-p) - 2 p = 0$
* $8 - 8p - 2p = 0$
* $10p = 8$
* $p = 0.8$
* (1) Estimated $p = 0.8$
* (2) Yes, $p > 0.6$
**b) Coin is Likely Fair (Prior around 0.5)**
* Prior: A beta distribution centered around 0.5. Let's use $\text{Beta}(5, 5)$, for example.
* $P(p) \propto p^4 (1-p)^4$
* Posterior: $P(p | \text{data}) \propto p^8 (1-p)^2 \cdot p^4 (1-p)^4 = p^{12} (1-p)^6$
* Maximize the posterior:
* $12 (1-p) - 6 p = 0$
* $12 - 12p - 6p = 0$
* $18p = 12$
* $p = 2/3 \approx 0.667$
* (1) Estimated $p \approx 0.667$
* (2) Yes, $p > 0.6$
**c) Coin is Likely Biased (Prior around 0 or 1)**
* Prior: A beta distribution centered around 0 or 1. Let's use $\text{Beta}(8, 2)$, for example.
* $P(p) \propto p^7 (1-p)^1$
* Posterior: $P(p | \text{data}) \propto p^8 (1-p)^2 \cdot p^7 (1-p)^1 = p^{15} (1-p)^3$
* Maximize the posterior:
* $15 (1-p) - 3 p = 0$
* $15 - 15p - 3p = 0$
* $18p = 15$
* $p = 5/6 \approx 0.833$
* (1) Estimated $p \approx 0.833$
* (2) Yes, $p > 0.6$
**d) 𝑝 can only take values 0.2, 0.7, or 0.9**
* Prior: We need to assign probabilities to each value. Let's assume equal prior probabilities: $P(0.2) = P(0.7) = P(0.9) = 1/3$.
* Likelihoods:
* $L(0.2) = 45 (0.2)^8 (0.8)^2 \approx 0.000046$
* $L(0.7) = 45 (0.7)^8 (0.3)^2 \approx 0.19$
* $L(0.9) = 45 (0.9)^8 (0.1)^2 \approx 0.196$
* Posterior (proportional to likelihood since prior is uniform):
* $P(0.2 | \text{data}) \propto 0.000046$
* $P(0.7 | \text{data}) \propto 0.19$
* $P(0.9 | \text{data}) \propto 0.196$
* The highest posterior is for $p = 0.9$.
* (1) Estimated $p = 0.9$
* (2) Yes, $p > 0.6$
**e) 0.4 ≤ 𝑝 ≤ 0.9**
* Prior: A uniform distribution on the interval $[0.4, 0.9]$.
* $P(p) \propto 1$ for $0.4 \le p \le 0.9$, and $0$ elsewhere.
* Posterior: $P(p | \text{data}) \propto p^8 (1-p)^2$ for $0.4 \le p \le 0.9$, and $0$ elsewhere.
* The maximum of $p^8 (1-p)^2$ occurs at $p = 0.8$, which is within the range $[0.4, 0.9]$.
* (1) Estimated $p = 0.8$
* (2) Yes, $p > 0.6$
**Summary**
In all cases, the estimated value of $p$ is greater than 0.6. The exact value of $p$ varies depending on the prior information, but it is always in the range of 0.667 to 0.9.
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