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?


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

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  • $\begingroup$ 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

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

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

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