STEM Problem Detail

Problem Statement

Prompt

Provide me the step by step solution of part a, b of question (1.2).

Solution Rendering...

**Solution:**

**(a) We have the following constraints:**

$$
X^{0}(p) = p_{1} + p_{2} + p_{3} + p_{4} + p_{5} + p_{6} - 1 = 0
$$

$$
X^{1}(p) = p_{3} - 2p_{2} = 0
$$

$$
X^{2}(p) = p_{4} - 2p_{5} = 0
$$

**We define**

$$
S^{*}(p,\lambda) = - \sum_{n} p_{n} \ln p_{n}
- \sum_{\alpha = 0}^{2} \lambda_{\alpha} X^{(\alpha)}(p),
$$

**and freely extremize over the probabilities**
$$\{p_{1},\ldots,p_{6}\}$$ **and the undetermined Lagrange multipliers**
$$\{\lambda_{0},\lambda_{1},\lambda_{2}\}$$.
**We obtain**

$$
\frac{\partial S^{*}}{\partial p_{1}} = -1 - \ln p_{1} - \lambda_{0}
$$

$$
\frac{\partial S^{*}}{\partial p_{2}} = -1 - \ln p_{2} - \lambda_{0} + 2\lambda_{1}
$$

$$
\frac{\partial S^{*}}{\partial p_{3}} = -1 - \ln p_{3} - \lambda_{0} - \lambda_{1}
$$

$$
\frac{\partial S^{*}}{\partial p_{4}} = -1 - \ln p_{4} - \lambda_{0} - \lambda_{2}
$$

$$
\frac{\partial S^{*}}{\partial p_{5}} = -1 - \ln p_{5} - \lambda_{0} + 2\lambda_{2}
$$

$$
\frac{\partial S^{*}}{\partial p_{6}} = -1 - \ln p_{6} - \lambda_{0}
$$

**Extremizing with respect to the undetermined multipliers generates the three constraint equations. We therefore have**

$$
p_{1} = e^{-\lambda_{0}-1}
$$

$$
p_{2} = e^{-\lambda_{0}-1} e^{2\lambda_{1}}
$$

$$
p_{3} = e^{-\lambda_{0}-1} e^{-\lambda_{1}}
$$

$$
p_{4} = e^{-\lambda_{0}-1} e^{-\lambda_{2}}
$$

$$
p_{5} = e^{-\lambda_{0}-1} e^{2\lambda_{2}}
$$

$$
p_{6} = e^{-\lambda_{0}-1}
$$

**We solve for** $$\{\lambda_{0},\lambda_{1},\lambda_{2}\}$$ **by imposing the three constraints. Let**

$$
x = p_{1} = p_{6} = e^{-\lambda_{0}-1}.
$$

**Then**

$$
p_{2} = x e^{2\lambda_{1}}, \quad
p_{3} = x e^{-\lambda_{1}}, \quad
p_{4} = x e^{-\lambda_{2}}, \quad
p_{5} = x e^{2\lambda_{2}}.
$$

**We then have**

$$
p_{3} = 2p_{2} \quad \Rightarrow \quad e^{-3\lambda_{1}} = 2
$$

$$
p_{4} = 2p_{5} \quad \Rightarrow \quad e^{-3\lambda_{2}} = 2
$$

**We may now solve for** $$x$$:

$$
\sum_{n=1}^{6} p_{n} = (2 + 2^{1/3} + 2^{4/3}) x = 1
\quad \Rightarrow \quad
x = \frac{1}{2 + 3 \cdot 2^{1/3}}.
$$

**We now have all the probabilities:**

$$
p_{1} = x = 0.1730
$$

$$
p_{2} = 2^{-2/3} x = 0.1090
$$

$$
p_{3} = 2^{1/3} x = 0.2180
$$

$$
p_{4} = 2^{1/3} x = 0.2180
$$

$$
p_{5} = 2^{-2/3} x = 0.1090
$$

$$
p_{6} = x = 0.1730
$$

**(b) The probability to roll a seven with two of these dice is**

$$
P(7) = 2p_{1}p_{6} + 2p_{2}p_{5} + 2p_{3}p_{4}
$$

$$
= 2\left(1 + 2^{-4/3} + 2^{2/3}\right) x^{2} = 0.1787.
$$

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Problem #24