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Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

$Y_2=α+βY_1+U \;where \; Y_1 \;and \;U\;is\;independent\;and\;U∼\mathcal{N}(0,v^2)$

Let $μ=350$ and $σ^2 =12365$

How do i find the distribution of $Y_2$?

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3 Answers 3

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If you like, define $Y_3 = \alpha + \beta Y_1$. Use properties of expectation. If $E[X] = \mu,$ then because expectation is a linear sum, $E[bX] = b\mu$, and $E[X+c] = \mu + c$ For example, if you double every number, the average will double, and if you add $5$ to every single number, the average goes up by $5$ as well. Using those, you can find that $Y_3$ is distributed $N(\alpha + \beta \mu, \beta^2 \sigma^2$)

Now you have a sum $Y_2 = Y_3 + U$ of independent variables. The mean of a sum is the sum of the means. The variance of the sum is the sum of the variances for independent variables. In formulas, $$E[X+Y] = E[X] + E[Y]$$ and $$V[X+Y] = V[X] + V[Y] + Cov[X,Y]$$. and for independent variables, the covariance is zero. Remember that standard deviation is the square root of the variance.

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  • $\begingroup$ Not quite sure i understand this method $\endgroup$
    – Marcus F
    Commented Mar 19, 2021 at 18:55
  • $\begingroup$ I added more detail for you. $\endgroup$ Commented Mar 19, 2021 at 20:27
  • $\begingroup$ So what would the mean and variance be in this case? $\endgroup$
    – Marcus F
    Commented Mar 19, 2021 at 22:56
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Using properties of Gaussians, Expectation and variance you get that

$$Y_2\sim N(\alpha+\beta\mu;\beta^2\sigma^2+v^2)$$

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  • $\begingroup$ Im not sure i understand this method $\endgroup$
    – Marcus F
    Commented Mar 20, 2021 at 17:08
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The CDF of random variable $Y_2$ is given by $$ F_{Y_2}(y_2)\equiv P(Y_2\leq y_2) $$ Therefore, $$ P(Y_2\leq y_2)=Pr(\alpha+\beta Y_1+U\leq y_2)=Pr(\beta Y_1+U\leq y_2-\alpha)=F_{\beta Y_1+U}(y_2-\alpha). $$ The sum of normal distributions is normal. Therefore, $\beta Y_1+U\sim N(\beta \mu,\beta^2\sigma^2+v^2)$.

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