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Suppose we have a random variable $X \sim Expo(\lambda)$ with support $X \in \mathbb{R^+}$.

$$f_X(x) =\lambda e^{-\lambda x}$$

Let us now also assume that $\lambda$ is itself a random variable and thus $(X,\lambda)$. Assume that it is also exponentially distributed such that $\lambda \sim Expo(\Lambda=2) $. Thus, we have the following PDF

$$f_\lambda(\lambda) =2e^{-2\lambda}$$

I have three related questions:

First, how do we you express the joint (PDF) distribution of both these random variables? Is the following the correct answer:

$$f_{X\lambda}(x,\lambda) = f_{X}(x)f_{\lambda}(\lambda) = \\ (2e^{-2\lambda}) (\lambda e^{-\lambda x}) $$

Second, how do we you express the joint (CDF) probability of both these random variables? Is the following correct?

$$F_{X\lambda}(x,\lambda) = F_{X}(x)F_{\lambda}(\lambda) = \\ \int_{0}^{\infty}\int_{0}^{\lambda} (2e^{-2\lambda}) (\lambda e^{-\lambda x})d\lambda dx $$

Here we first integrate over $\lambda$ since $X$ is conditioned on it. And then we integrate over $X$?

Third, how do we calculate $\mathbb{E}[X|\lambda]$?

$$ \mathbb{E}[X|\lambda] = \int_{0}^{\infty}\int_{0}^{\lambda} \lambda (2e^{-2\lambda}) x(\lambda e^{-\lambda x})d\lambda dx $$

Here, as in the second question, we have to integrate over $\lambda$ since $X$ is conditioned on it.

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First, how do we you express the joint (PDF) distribution of both these random variables? Is the following the correct answer:

I'd say $$f_{X\lambda}(x,\lambda) = f_{X \mid \lambda}(x \mid \lambda)f_{\lambda}(\lambda) = \\ (2e^{-2\lambda}) (\lambda e^{-\lambda x})$$.

Second, how do we you express the joint (CDF) probability of both these random variables? Is the following correct?

You can only split up probabilities like that if the two random variables are independent, which they are not. Take the joint and integrate over the right rectangle. \begin{align*} F_{X\lambda}(x,l) &= P(X \le x, \lambda < l) \\ &= \int_{0}^{x}\int_{0}^{l} (2e^{-2\lambda}) (\lambda e^{-\lambda s})d\lambda ds. \end{align*}

Third, how do we calculate $\mathbb{E}[X \mid \lambda]$?

You just use the conditional density $f_{X \mid \lambda}(x \mid \lambda)$ and integrate: $$ \mathbb{E}[X \mid \lambda] = \int_0^{\infty}xf_{X \mid \lambda}(x \mid \lambda)dx = \int_0^{\infty}x \lambda e^{-\lambda x} = \lambda^{-1}. $$

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