-4
$\begingroup$

If $X$ and $Y$ are independent random variables, with respective moment-generating functions (mgf) $M_X(t)$ and $M_Y(t)$, the mgf of $2X+Y$ is $M_X(2t)M_Y(t)$. What would the mgf of $2X-Y$ be?

$\endgroup$

closed as off-topic by Em., Did, zoli, Daniel W. Farlow, C. Falcon May 14 '17 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Em., Did, zoli, Daniel W. Farlow, C. Falcon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ To the people down-voting: why? The text I am working from has a very sparse description of mgf and I am trying to better understand their logic from the few examples given. I try to avoid asking bad or non-useful questions so let me know what's wrong. $\endgroup$ – user1569317 May 13 '17 at 21:44
1
$\begingroup$

Since $X$ and $Y$ are independent, we have $E(e^{aX}e^{bY})=E(e^{aX})E(e^{bY})$. Thus, we have: $$M_{aX+bY}(t)=E\left(e^{(aX+bY)t}\right)=E\left(e^{(aX)t}e^{(bY)t}\right)=E\left(e^{X(at)}\right)E\left(e^{Y(bt)}\right)=M_{X}(at)M_Y(bt)$$

$\endgroup$
  • $\begingroup$ @PMF this answer still works: use $a=2$ and $b=-1$. $\endgroup$ – Dave May 14 '17 at 14:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.