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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!

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    $\begingroup$ Unfortunately $P(Y_1=2)=0$, so the last formula won't be helpful here. Also note that $Y_1$ and $Y_2$ are not independent (not that you said they were, but just a pitfall warning). Personally, I would subtract $1$ from $X_1$ and $X_3$ and rephrase the problem, so that all random variables are centered at $0$. What happens if you condition on $Y_1=2,X_1=t$ and then integrate over $t$? $\endgroup$ Commented Jul 9, 2020 at 4:17
  • $\begingroup$ The command of Mathematica 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 $\frac{1}{2} \left(1-\text{erf}\left(\frac{3}{2 \sqrt{115}}\right)\right)=0.421595$. $\endgroup$
    – user64494
    Commented Jul 9, 2020 at 4:31
  • $\begingroup$ It can be shown in this way that $(Y_2 \mid Y_1 = 2) \overset d = Y_2 - 2 (Y_1 - 2)/5$. $\endgroup$
    – Maxim
    Commented Jul 9, 2020 at 9:56

1 Answer 1

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Note that $Y_1$ and $Y_2$ are jointly Gaussian, so we can use the following fact of jointly Gaussian variables $Y_1 \sim \mathcal{N}(\mu_1, \sigma^2_1)$ and $Y_2 \sim \mathcal{N}(\mu_2, \sigma^2_2)$: $$Y_2 ~|~ (Y_1 = y) \sim \mathcal{N}\left(\mu_2 + \rho \cdot\frac{\sigma_2}{\sigma_1} \cdot (y - \mu_1), (1 - \rho^2) \cdot\sigma^2_2\right)$$ where $\rho$ is the correlation coefficient between $Y_1$ and $Y_2$: $$\rho = \frac{\text{Cov}(Y_1, Y_2)}{\sigma_1 \sigma_2}$$ In this case, $Y_1 \sim \mathcal{N}(1, 5)$ and $Y_2 \sim \mathcal{N}(3, 10)$ (note this is not exactly what you've written; when multiplying a normal distribution by a constant $c$ the variance is scaled by $c^2$, not $c$), so $$\text{Cov}(Y_1, Y_2) = \text{Cov}(X_1, X_2) + \text{Cov}(X_1, 3 X_3) + \text{Cov}(2 X_2, X_2) + \text{Cov}(2 X_2, X_3)$$ and all terms vanish except for the third, which evaluates to $2 \cdot \text{Cov}(X_2, X_2) = 2 \cdot \text{Var}(X_2) = 2$. Thus, $\rho = 2 / \sqrt{50}$. Plugging in our formula, the conditional distribution is then given by $$Y_2 ~|~ (Y_1 = 2) \sim \mathcal{N}\left(3 + \frac{2}{\sqrt{50}} \cdot \frac{\sqrt{17}}{\sqrt{5}}, \frac{23}{25} \cdot 10 \right) = \mathcal{N}\left(\frac{17}{5}, \frac{46}{5} \right)$$ So it follows that $$\mathbb{P}(Y_2 > 4 ~|~ Y_1 = 2) = 1 - \Phi\left(\frac{3/5}{\sqrt{46 / 5}}\right) = \boxed{1 - \Phi\left(\frac{3}{\sqrt{230}}\right)}$$ which matches up with the WA answer in the comments! Thanks to Maxim for pointing out the error.

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  • $\begingroup$ We want to compute the distribution of $Y_2 \mid Y_1 = 2$, not $Y_1 \mid Y_2 = 2$. $\endgroup$
    – Maxim
    Commented Jul 9, 2020 at 10:02
  • $\begingroup$ Ah thanks so much @Maxim! $\endgroup$
    – paulinho
    Commented Jul 9, 2020 at 14:11

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