Sum of indepedent random variables and a constant

Question

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.

Answer 1

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 $\sigma_Y=\sqrt{\sigma_1^2+\sigma_2^2}, \mu_Y=c+\mu_2-\mu_1$.