What Is the MGF of X − Y?

The moment-generating function (MGF) of a random variable $Z$ is $M_Z(t)=\mathbb{E}[e^{tZ}]$, for values of $t$ where the expectation is finite. To find the MGF of a difference, the key step is to keep track of the minus sign on $Y$.

If $X$ and $Y$ are independent, then

The formula is valid on the range of $t$ where both factors on the right are finite. If $X$ and $Y$ are dependent, this product rule does not generally apply; use the joint distribution to calculate $\mathbb{E}[e^{t(X-Y)}]$.

Deriving the formula

Start from the definition and rewrite the exponential:

When $X$ and $Y$ are independent, functions of $X$ and $Y$ are independent as well. Therefore the expectation factors:

The argument of $M_Y$ is $-t$ because the exponent contains $-tY$. This is the main detail that distinguishes the MGF of a difference from the MGF of a sum. For independent variables, the sum formula is $M_{X+Y}(t)=M_X(t)M_Y(t)$; for a difference, replace the second $t$ with $-t$.

Worked example: independent exponential variables

Suppose $X$ and $Y$ are independent exponential random variables with rates $\lambda>0$ and $\mu>0$. An exponential variable with rate $r$ has MGF

Thus $M_X(t)=\frac{\lambda}{\lambda-t}$ for $t<\lambda$, while

The condition $-t<\mu$ is equivalent to $t>-\mu$. Combining the two domain conditions gives $-\mu<t<\lambda$. Consequently,

For example, if both rates are $2$, then

This MGF is symmetric in $t$, so its first derivative at zero is zero. That matches $\mathbb{E}[X-Y]=\mathbb{E}[X]-\mathbb{E}[Y]=\frac12-\frac12=0$. Its second derivative at zero is $\frac12$, which is the variance of $X-Y$: independence gives $\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)=\frac14+\frac14=\frac12$.

If X and Y are dependent

Without independence, the correct starting point remains

but you cannot usually split this into a product of marginal expectations. If $(X,Y)$ has a joint density $f_{X,Y}(x,y)$, for instance, use

provided the integral is finite. A joint probability mass function gives the analogous sum. Dependence matters because knowing the separate distributions of $X$ and $Y$ does not, in general, tell you the distribution of their difference.

For example, if $Y=X$, then $X-Y=0$ with probability one, so $M_{X-Y}(t)=1$. Multiplying $M_X(t)$ by $M_Y(-t)$ would generally give a different answer: that product treats the variables as independent, not identical copies that are equal on every outcome.

Common mistakes

• Using $M_Y(t)$ instead of $M_Y(-t)$. The negative argument follows directly from $e^{t(X-Y)}=e^{tX}e^{-tY}$.
• Multiplying MGFs without independence. The factorization is justified by independence, not merely by knowing both marginal MGFs.
• Ignoring the domain. In the independent case, $t$ must belong to the domain of $M_X(t)$ and also make $M_Y(-t)$ finite. The example’s domain is the intersection $(-\mu,\lambda)$.
• Confusing a difference with a ratio. The MGF concerns $e^{t(X-Y)}$; it does not directly give the distribution or MGF of $X/Y$.

Once the MGF is known near zero, derivatives can provide moments when they exist. In particular, for independent variables with finite means and variances, $\mathbb{E}[X-Y]=\mathbb{E}[X]-\mathbb{E}[Y]$ and $\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$.