# Transformation of a normal distribution- Find marginal distribution

I got these 2 models for hourly wage for 2 periods:

The hourly wage for period 1 is normally distributed with mean $$µ$$ and variance $$σ^2$$ so $$Y_1 \sim N(\mu,\sigma^2)$$.

And the hourly wage for period 2 is given by: $$Y_2=\alpha+\beta Y_1+U$$ where $$Y_1$$ and $$U$$ are independent and $$U \sim N(0,v^2)$$. Then we assume that $$\beta \neq 0$$ and let $$\mu = 350$$ and $$\sigma^2=12365$$ and $$\alpha=350\cdot(1-\beta)$$ and $$v^2=12365 \cdot (1-\beta^2)$$.

Now I have to find the marginal distributions of $$Y_1$$ and $$Y_2$$. But how can I found that? Normally I would integrate by $$Y_1$$ to find the marginal distribution of $$Y_2$$ and integrate by $$Y_2$$ to find the marginal distribution of $$Y_1$$. But here I dont have an expression with both $$Y_1$$ and $$Y_2$$?

No integrals are needed. Linear combinations of gaussians are still gaussian thus all you have to do is to calculate mean and variance of $$Y_2$$ and I explained you how to fo that in my answer to your previous question
• Yes now I have found that $Y_2$ is distributed by $N(\alpha+\beta \cdot \mu,\beta \cdot \sigma^2+v^2)$. So now I have to plot in the new expressions of $\mu$, $\sigma^2$, $\alpha$ and $v^2$? Or what do I have to do? Mar 20, 2021 at 15:15
• I have found that on wikipedia that the marginal distribution is given by $p_{X}(x)=E_Y[P_{X|Y}(x|y)]$. And I have earlier found that $E(Y_2|Y_1)=\alpha+\beta \cdot Y_1$. Is that the same and can I use this? Mar 20, 2021 at 16:44