Let $X\sim N(12,4)$ and $Y \sim N(3,1)$

Let $Z = X - Y$

Find the Moment Generating Function of $Z$.

I tried finding the expected value of $e$ to the power of $tz$, but this isn't possible to separate in the expected value function. I know how to use find the MGF when it is a sum of 2 random variables, but what is the technique when it is a difference like this?

  • 3
    $\begingroup$ Use $-Y \sim N(-3,1)$ and add. $\endgroup$
    – whuber
    Jan 20, 2020 at 20:52
  • $\begingroup$ Why is the variance not also negative? $\endgroup$
    – Spencer
    Jan 20, 2020 at 21:26
  • 1
    $\begingroup$ There's no such thing as a negative variance. Negating a variable doesn't change the amount it varies around its mean, so the variance remains the same. $\endgroup$
    – whuber
    Jan 21, 2020 at 0:25
  • $\begingroup$ @whuber Nowhere does the OP say that the random variables are jointly normal $\endgroup$ Feb 15, 2021 at 0:17
  • $\begingroup$ @Dilip That is true and well worth remembering. But that point has been made here on CV ad nauseam, so it would suffice for a respondent to make the assumption explicit and move on. $\endgroup$
    – whuber
    Feb 15, 2021 at 13:35

1 Answer 1


Continuing from @whuber's comment, $-Y$ has normal distribution with mean $-3$ and variance $1$. So $Z = X - Y = X + (-Y)$ has normal distribution with mean $12-3=9$ and variance $4+1=5$. The moment generating function of a normal distribution with mean $\mu$ and variance $\sigma^2$ is $e^{\mu t + \sigma^2 t^2/2}$, and so the moment generating function of $Z$ is $$ e^{9t + \frac52 t^2}. $$


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