# Sum of indepedent random variables and a constant

Let $$X_1$$ and $$X_2$$ be independent Normal random variables with mean $$\mu_1$$ and $$\mu_2$$, and variances $$\sigma_1$$ and $$\sigma_2$$. Let $$Y = X_2-X_1 + c$$, where $$c$$ is a constant.

For notational simplicity, let $$Y = X' + c$$, where $$X' = X_2-X_1$$

The moment generating function of a Normal random variable $$X_n\sim N(μ_n,σ_n^2)$$ is $$M_{X_n } (t)=\text{exp}\Big(μ_n t+\frac{σ_n^2 t^2}{2}\Big)$$

Then, $$M_{X'} (t)=\text{exp}\Big[t(\mu_1+\mu_2)+\frac{t^2}{2}(\sigma^2_1+\sigma^2_2)\Big]$$

Thus, by the uniqueness property of moment generating function, $$X'$$ is said to follow a Normal distribution with mean $$(\mu_1+\mu_2)$$ and variance $$(\sigma^2_1+\sigma^2_2)$$.

How do I proceed from here to derive the following:

1. First, I want to establish that $$Y$$ is also a Normal random variable using the moment generating function. The constant $$c$$ bugs me, but I cannot ignore it. As per my understanding, $$Y\sim N\big(c+\mu_1+\mu_2,\sigma^2_1+\sigma^2_2\big)$$ (and I could be completely wrong). But I don't understand how to apply the moment generating function to prove that.

2. Secondly, I want to derive $$P(Y \leq a)$$, where $$a$$ is a constant. If I am not mistaken, $$P(Y \leq a)$$ can be written as $$P(Y \leq a)=\Phi\Bigg(\frac{a-X_2-X_1 + c}{\sqrt{2(\sigma^2_1+\sigma^2_2)}}\Bigg).$$ But how do I arrive at that.

Any help is high solicited and appreciated.

It should be $$\mu_2-\mu_1$$ everywhere, not $$\mu_2+\mu_1$$. For the constant, you have $$M_c(t)=e^{ct}$$ and when multiplied with $$M_{X'}(t)$$, you have $$M_Y(t)=\exp(t(c-\mu_1+\mu_2)+...)$$ So, by uniqueness property, you show $$Y$$ is normal.
For the second, $$P(Y\leq a)=P\left(\underbrace{\frac{Y- \mu_Y}{\sigma_Y}}_{\text{ Standard normal}}\leq \frac{a-\mu_Y}{\sigma_Y}\right)=P\left(Z\leq\frac{a-\mu_Y}{\sigma_Y}\right)=\Phi\left(\frac{a-\mu_Y}{\sigma_Y}\right)$$
where $$\sigma_Y=\sqrt{\sigma_1^2+\sigma_2^2}, \mu_Y=c+\mu_2-\mu_1$$.