To find α and β^2, we can use Bayes' theorem and apply the formula for the posterior distribution of θ given the observed data.
By Bayes' theorem, we have:
π(θ|X₁,...,Xₙ) ∝ p(X₁,...,Xₙ|θ)π(θ)
Since the random variables X₁,...,Xₙ are i.i.d. and conditional on θ, the joint density of X₁,...,Xₙ given θ is the product of their individual conditional densities:
p(X₁,...,Xₙ|θ) = p(X₁|θ)p(X₂|θ)...p(Xₙ|θ)
Substituting the given conditional density of Xᵢ given θ, we have:
p(X₁,...,Xₙ|θ) = N(X₁|θ,1)N(X₂|θ,1)...N(Xₙ|θ,1)
Expanding the product, we have:
p(X₁,...,Xₙ|θ) = N(X₁|θ,1)N(X₂|θ,1)...N(Xₙ|θ,1)
= (2π)⁽⁻ⁿ/²⁾exp⁽⁻½[(X₁-θ)²+(X₂-θ)²+...+(Xₙ-θ)²]⁾
Next, we can substitute the prior distribution of θ into Bayes' theorem:
π(θ|X₁,...,Xₙ) ∝ p(X₁,...,Xₙ|θ)π(θ)
= (2π)⁽⁻ⁿ/²⁾exp⁽⁻½[(X₁-θ)²+(X₂-θ)²+...+(Xₙ-θ)²]⁾ * N(θ|μ,1)
Taking the logarithm of both sides, we can simplify the expression:
log(π(θ|X₁,...,Xₙ)) = -½[(X₁-θ)²+(X₂-θ)²+...+(Xₙ-θ)²] + log(N(θ|μ,1)) + C
Where C is a constant not dependent on θ.
Expanding the terms inside the square brackets, we have:
log(π(θ|X₁,...,Xₙ)) = -½[X₁²+θ²-2X₁θ+X₂²+θ²-2X₂θ+...+Xₙ²+θ²-2Xₙθ] + log(N(θ|μ,1)) + C
= -½[nθ²+X₁²+X₂²+...+Xₙ²-2θ(X₁+X₂+...+Xₙ)] + log(N(θ|μ,1)) + C
= -½(nθ²-2θ(X₁+X₂+...+Xₙ) + (X₁²+X₂²+...+Xₙ²)) + log(N(θ|μ,1)) + C
= -½(nθ²-2θ(X₁+X₂+...+Xₙ) + Σ(Xᵢ)²) + log(N(θ|μ,1)) + C
Comparing this expression with the standard form of a Gaussian distribution N(α,β²), we can determine the values of α and β². By matching the terms, we can see that:
α = (X₁+X₂+...+Xₙ)/n + μ/(n+1)
β² = 1/(n+1)
Therefore, the correct expressions for α and β² are:
α = (1/n)(ΣXᵢ) + μ/(n+1)
β² = 1/(n+1)
Let X_1,X_2,\dots ,X_ n be random variables; which are i.i.d., conditional on \theta, and such that,
p(X_ i|\theta )=N(\theta ,1),
where p(X_ i|\theta ) is the conditonal density of X_ i given \theta. Furthermore, we assume the prior \pi (\theta )\sim N(\mu ,1). Let
\pi (\theta |X_1,\dots ,X_ n)\sim N(\alpha ,\beta ^2).
Find, \alpha and \beta ^2.
\alpha =
\frac{1}{n+1}\left(\left(\sum _{i=1}^ n X_ i\right)+\mu \right)
\frac{1}{n}\left(\sum _{i=1}^ n X_ i+\mu \right)
\frac{1}{n+1}\left(\sum _{i=1}^ n X_ i\right)+\mu
\frac{1}{n}\left(\sum _{i=1}^ n X_ i\right)
unanswered
\beta ^2=
\frac{1}{n+1}
\frac{1}{n}
\frac{\mu }{n}
\frac{\mu }{n+1}
1 answer