# Are $5 \text{ x } m_y(s)$ and $m_y(s)^2$ moment generating functions

Say if I had some MGF like $m_y(s)$ of some random variable $Y$, are $5 \text{ x } m_y(s)$ and $m_y(s)^2$ moment generating functions? This is a curiosity thing, does multiplying through by some integer have an effect? How about squaring the MGF?

In general no, a scalar multiple of a Moment Generating Function (MGF) or a square of an MGF will not be a MGF. Consider the MGF of a standrd normal density

$M_X(t)=e^{t^2/2}.$

Then the scalar multiple in the post would be

$5e^{t^2/2}$ which is not an MGF.

If you multiply your random variable by a constant then the resulting MGF of that random variable $Y=5X$ (say) would be

$\int e^{t5x}f(x)dx = \int e^{t^\prime x}f(x)dx = M_X(5t) \neq 5M_X(t)$ unless of course your MGF was a linear function in $t$.

• Would it be the same for a square, i.e $[e^{\frac{t^2}{2}}]^2$ is not an MGF? – Rubicon Aug 18 '17 at 3:04
• The argument is not the same because a square is a different sort of function. The answer is still the same though that in general the function may not be an MGF. You can find examples where it will be an MGF but in general it will not be an MGF. – Lucas Roberts Aug 18 '17 at 3:08