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I have two continuous random variables $V_1$ and $V_2$ defined as

$$\begin{aligned}V_1 &:= a_1 \cdot W_1 + a_2 \cdot W_2 + a_3 \cdot W_3 + a_4 \cdot W_4 + a_5 \cdot W_5 \\ V_2 &:= b_1 \cdot Y + b_2 \cdot W_2 + b_3 \cdot W_3 + b_4 \cdot W_4 + b_5 \cdot W_5\end{aligned}$$

where $W_1$, $W_2$, $W_3$, $W_4$, $W_5$ and $Y$ are mutually independent continuous random variables with known Gaussian distributions. Could anyone please help me with the methodology of finding a conditional probability density function $p(V_1|V_2)$?

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Let $\{W_i\}$ be set of $W_i$ for $i=2,3,4,5$.

$V_1$ and $V_2$ are not independent. But there are conditionally independent given $\{W_i\}$!. Therefore:

$$p(V_1, V_2 \lvert \{W_i\}) = p(V_1 \lvert \{W_i\}) p(V_2 \lvert \{W_i\})$$

From law of total probability:

$$p(V_1, V_2 ) = \int p(V_1, V_2 \lvert \{W_i\}) p( \{ W_i \} )$$

Then you can easily compute $p(V_2)$ since $\{W_i\}$ are mutually independent Gaussians, so sum of them is another Gaussian, $\sum W_i = N(\sum \mu_i, \sum \sigma^2_i)$, where $\mu_i, \sigma^2_i$ are means and variances of $W_i$.

Then, from definition: $$p(V_1 \lvert V_2) = p(V_1, V_2) /p(V_2)$$

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  • $\begingroup$ Thank you very much! $\endgroup$ – Nadazhda Mar 23 at 10:37

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