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Suppose you have some moment generating function $M_x(t)$ Now all the random variables x are increased by a arbitrary value b. What is the new moment generating value?

I tried solving this by moving back from the MGF to the probability distribution, but this proved impossible. Could somebody give me a nudge in the right direction? I'm looking for a 'proven' method instead of using intuition. Thanks!

Edit:

Is this correct?

Declare a new variable $y = x + b$, so $x = y - b$

$E(e^{tx}) = E(e^{ty-tb}) = E(e^{ty})E(e^{-tb}) = M_y(t)E(e^{-tb}) =M_y(t)e^{-tb}$

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  • $\begingroup$ Can you clarify? You use $x$ as argument to the mgf $M(x)$ and then name it all the random variables x, but the argument to the mgf is not the random variable! $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 25, 2020 at 13:56
  • $\begingroup$ Yes I get that. Let me rephrase it: I have some random variable x and the moment generating function of x is M(t) $\endgroup$
    – Tijmen
    Commented Sep 25, 2020 at 13:59
  • $\begingroup$ Dear @Tijmen , please, update your question. $\endgroup$
    – ABK
    Commented Sep 25, 2020 at 14:15
  • $\begingroup$ I already did update the question $\endgroup$
    – Tijmen
    Commented Sep 25, 2020 at 14:17
  • $\begingroup$ Yes, that is correct! Now you can answer your own question, formally, so it does not linger as unanswered. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 25, 2020 at 14:29

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$\begingroup$

Declare a new variable $y = x + b$, so $x = y - b$

$E(e^{tx}) = E(e^{ty-tb}) = E(e^{ty})E(e^{-tb}) = M_y(t)E(e^{-tb}) =M_y(t)e^{-tb}$

Improve this answer
$\endgroup$

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