Let $X_1, X_2$ and $X_3$ be independent random variables, such that $X_1$ and $X_3$ are distributed $N (1, 1)$ and $X_2$ $N (0, 1)$ is distributed. Define $Y_1 = X_1 + 2X_2$ and $Y_2 = X_2 + 3X_3$. Determine $P(Y_2> 4 | Y_1 = 2)$. Express the result in terms of $\Phi$, the cumulative probability function of the standard normal distribution.
My try:
I know that if $U_i ∼ N (µ_i, {σ^2}_i), i = 1, 2$, independent random varaibles. Then $U_1 + U_2 ∼ N (µ_1 + µ_2, {σ^2}_1 + {σ^2}_2).$
Then $$Y_1 = X_1 + 2X_2 = X_1 + X_2 + X_2∼N(1,3) $$ $$Y_2 = X_2 + 3X_3 = X_2 + X_3 + X_3 ∼N(3,4)$$
$P(Y_2> 4 | Y_1 = 2) = 1 - P(Y_2 \leq 4 | Y_1 = 2) = 1 - \frac{P(Y_2 \leq 4 , Y_1 = 2)}{P(Y_1=2)}$
I don´t know If I´m on the right track. Any suggestions would be great!
Probability[ x2 + 3 x3 > 4 \[Conditioned] x1 + 2*x2 == 2, {x1 \[Distributed] NormalDistribution[1, 1], x3 \[Distributed] NormalDistribution[1, 1], x2 \[Distributed] NormalDistribution[0, 1]}]
answers $\frac{1}{2} \left(1-\text{erf}\left(\frac{3}{2 \sqrt{115}}\right)\right)=0.421595$. $\endgroup$