A simple baysian question on likelihood

[ghostyc]

Hi all,

In this question, I found that

$$\Pr(X_i|\theta)=\frac{\exp(\theta x_i)}{1+\exp(\theta)}$$

and I carry on with the likelihood being $$\frac{\exp(\theta \sum x_i)}{(1+\exp(\theta))^n}$$

and so $$s=\sum x_i = Tn$$

I need some help with part (c).

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How do I start with this question then?

I couldn't get a general expression for $$\Pr(X_i|\theta)$$.


How do I generally deal with these "piecewise probability density function"?

Thanks!

 

[Stephen Tashi]

Suppose that $X_1,...,X_n$ are independent and identically distributed conditional on a random parameter $\theta > 0$ such that
$$P(X_i = 1|\theta) = \frac{e^\theta}{1+e^\theta}$$ and $$P(X_i = 0| \theta) = 1 - P(X_i=1| \theta)$$.

A normal prior for $\theta$ is taken with mean $\mu$ and variance $\sigma^2$.

(a) Show that the liklihood function is of this type

$$l(\theta) = \frac{e^{\theta S}}{(1 + e^\theta)^n} = \big{(}\frac{e^{\theta T}}{1 + e^\theta}\big{)}^n$$ and find $S$ and $T$ in terms of $\{X_1,X_2,...X_n\}$.

(b) Hence, write down the posterior distribution for $\theta$.

(c) For any $0 < t < 1$ and $\theta > 0$, show that

$$\frac{e^{\theta t }}{1 + e^\theta} < t^t(1-t)^{1-t}$$ .

In this question, I found that

$$\Pr(X_i|\theta)=\frac{\exp(\theta x_i)}{1+\exp(\theta)}$$

and I carry on with the likelihood being $$\frac{\exp(\theta \sum x_i)}{(1+\exp(\theta))^n}$$

and so $$s=\sum x_i = Tn$$

I need some help with part (c).

One thought that comes to mind is to use calculus to see if the function
$$f(\theta) = (\theta)^S ( 1 - \theta)^{n-S}$$ takes its maximum value when $$\theta = S/N$$. If so, then $$(f(\theta))^{1/n}$$ also takes its maximum value there.

 

[Stephen Tashi]

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Suppose that $X_1,...,X_n$ are independent and identically distributed conditional on a random parameter $\theta > 0$ such that
for each $i = 1,2,...n$

$$P(X_i = -1|\theta) = \theta}$$ and $$P(X_i = 1| \theta) = 1 - \theta$$.

(a) Show that the liklihood function is given by the form

$$l(\theta) = \theta^{n/2 - S} (1-\theta)^{n/2 + S}$$ and find $S$ in terms of $\{X_1,X_2,...X_n\}$.
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let [itex] k [/tex] be the number of the $X_i$ that are +1 and let $S = n/2 - k$

Isn't "discrete" a better term than "piecewise"?