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There is a question that how to draw the estimators (ratio estimator) if we know the prior for the numerator or the ratio estimator.

Assume that we have three coefficients: $a_1,a_2$ and $a_3$ . And we beleive that all the three coefficients follow this rule:

$a_i=1-F(i)$ while $F(i)=e^{-\lambda*i}$. $F(i)$ is the cdf of the exponential distributions.

There are some restriction for the three coefficients:$1>a_1>a_2>a_3>0$.

Now the story is: we have a dataset $Z_t$ which is a survival dataset or time series data that can be used to estiamte the three coefficients. Let's assume that we can estimate them and obtain the prior: $\hat{a}_i=1-\hat{F}(i)=e^{-\hat{\lambda}*i}$ . When we estimate those coefficients, we use the data which is discrete time data. For example, i=1,2, and 3. So we just need to estimate $a_i$ with $i=1,2,3$ by using the discrete time data. So $\hat{a}_1,\hat{a}_2,\hat{a}_3$ are estimators and each of them has their own variance (you can thing $\hat{a}_1$ with i=1,2,3 follow normal distributions asymptotically with mean=E(1-F(i)), $Var=\sigma_i^2$. We can also use the discrete time datasect $Z_t$ to calculate $\sigma_i^2$ with i=1,2,3)

After that, we define some new estimators say $\hat{\beta}_i=\frac{\hat{a}_i}{\hat{a}_1+\hat{a}_2+\hat{a}_3}$ with i=1,2,3. We have another four time series dataset:$Y_{t}, X_{1,t},X_{2,t},X_{3,t}$.

$Z_t$ is different from $Y_{t}, X_{1,t},X_{2,t},X_{3,t}$.

We also assume that there is a theorem (we call it as equation (1.1)):

$Y_t=\beta_1 X_{1,t}+\beta_2 X_{2,t}+\beta_3 X_{3,t}$

Where ${\beta}_1+{\beta}_2+{\beta}_3=1$ and $1>{\beta}_1>{\beta}_2>{\beta}_3>0$.

Our propuse is to draw $\hat{\beta}_{i,bayes}$ from equation (1.1) by appliyng the prior of $\hat{\beta}_i$. After that, we compare $\hat{\beta}_{i,bayes}$ with $\hat{\beta}_{i}$ which is directly calculated from the data $Z_t$. If equation (1,1) is correct, $\hat{\beta}_{i,bayes}$ should be close to $\hat{\beta}_{i}$.

Note that $\hat{\beta}_{i,bayes}$ must satisfy the two restrictions (sum to 1 and decrese with i).

That is my question. Any ideas?

For myself, I thought Dirchlet distribution may satisfy one condition (summation equals to 1). But Dirchlet distribution assume that $\hat{\beta}_{i,bayes}$ follows Gamma disribution. I also thought I can add some constraint to the bayesian method or use lasso method. But I think they may not work(?).

Mike
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  • Your conditions contradict each other. If $a_1$ is exponentially distributed there is a positive chance that $a_1>1$. – kimchi lover Mar 08 '24 at 12:05
  • Thanks so much for your comment @kimchi lover. I correct my descriptions of the coefficient. They should be the CDF of exponential distribution. not pdf. – Mike Mar 09 '24 at 08:37
  • Should the first part of your question be like the following?$Z_t$ follows exponential distribution $Exp(\lambda)$. One can use the observed values of $Z_t$ to estimate $1/\lambda$ by the sample average of $Z_t$. Then one can construct $\hat{a}_i$ using the estimated CDF.Asymptotically, the sample average of $Z_t$, or $1/\hat{\lambda}$ should follow normal distribution. So you could study the distribution of $\hat{a}_i$. If you know $\hat{a}_i$ and calculate $\hat{\beta}_i$ in the way you described, the sum of $\hat{\beta}_i$ should be 1 and the order condition be satisfied. – Jack Mar 09 '24 at 11:12
  • Your $F$ is not a cdf: If $\lambda>0$ it is decreasing, if $\lambda<0$, it is unbounded. (And if $\lambda=0$, your $1>a_1>a_2$ fails.) – kimchi lover Mar 09 '24 at 12:01

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