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I'm trying to solve myself such a problem: calculation of Bayes Factor for two groups from exponential distributions. $$x_1, x_2, \ldots, x_n\sim \exp(\lambda_1)$$ $$y_1, y_2, \ldots, y_m\sim \exp(\lambda_2)$$

I will show what I have already done and my two questions are below.

The null hypothesis is $$H_0: \mu_1=\mu_2$$ but here it translates to $$H_0:\lambda_1=\lambda_2$$ I need two entities: likelihood function for this data and priors on lambdas. The likelihood seems to be: $$L(\lambda_1, \lambda_2\mid x_1, \ldots, x_n, y_1\ldots, y_m)= \lambda_1^n\cdot \exp(-\lambda_1\cdot n\cdot\bar{x}) \cdot\lambda_2^m\cdot\exp(-\lambda_2\cdot m\cdot\bar{y}) $$

To calculate Bayes Factor, I need to have two marginal likelihoods: $$m_0(x_1, \ldots, x_n,y_1, y_2, \ldots, y_m)$$ for null model ($\lambda_1=\lambda_2$), and $$m_1(x_1, \ldots, x_n,y_1, y_2, \ldots, y_m)$$ for alternative model ($\lambda_1\neq\lambda_2$).

Q1. Does those marginals look like this?: $$m_0(x_1, \ldots, x_n,y_1, y_2, \ldots, y_m)=\int \lambda^{n+m}\cdot \exp(-\lambda\cdot (n\cdot\bar{x} + m\cdot\bar{y}))\cdot \pi(\lambda) d\lambda$$

$$m_0(x_1, \ldots, x_n,y_1, y_2, \ldots, y_m)=\int \lambda_1^n\cdot \exp(-\lambda_1\cdot n\cdot\bar{x}) \cdot\lambda_2^m\cdot\exp(-\lambda_2\cdot m\cdot\bar{y})\cdot\pi(\lambda_1, \lambda_2)\ d\lambda_1 d\lambda_2$$

where $\pi(\cdot)$ means prior.

Q2. I guess, I need a prior $\pi(\lambda)$ on $\lambda$ for null model and prior $\pi(\lambda_1, \lambda_2)$ for $\lambda_1$ and $\lambda_2$ for alternative model?

Is it possible to assume that $\pi(\lambda_1, \lambda_2)=\pi(\lambda_1)\cdot \pi(\lambda_2)$ (independence of $\lambda_1$ and $\lambda_2$)?

Michael R. Chernick
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Lil'Lobster
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