# Distribution of continous random variable

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 $$Y_2$$?

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.

• Not quite sure i understand this method Commented Mar 19, 2021 at 18:55
• I added more detail for you. Commented Mar 19, 2021 at 20:27
• So what would the mean and variance be in this case? Commented Mar 19, 2021 at 22:56

Using properties of Gaussians, Expectation and variance you get that

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

• Im not sure i understand this method Commented Mar 20, 2021 at 17:08

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, $$\beta Y_1+U\sim N(\beta \mu,\beta^2\sigma^2+v^2)$$.