The following question is from my book:

If the mgf of $X$ is

$M(t) = \dfrac {e^{5t} - e^{4t}}{t}, t \not = 0$, and $M(0)=1$,

find (a) $E(X)$, (b) Var$(X)$, and (c) $P(4.2 < X \le 4.7)$

I that to find the mean and the variance I just compute $M'(0)$ and $M''(0)-[M'(0)]^2$. But I'm not sure how to do part c. Am I supposed to guess the PDF $f(x)$, knowing that $\displaystyle \dfrac {e^{5t} - e^{4t}}{t} = \int_{-\infty}^{\infty} e^{tx} f(x)dx?$

  • 1
    $\begingroup$ can you add the self-study tag? $\endgroup$ – kjetil b halvorsen Jun 8 '17 at 13:52
  • $\begingroup$ This should not be a guess. The question should provide enough information to derive the pdf. from the properties given for the mgf. $\endgroup$ – Michael R. Chernick Jun 8 '17 at 14:00
  • $\begingroup$ Here is a hint: the probability you are looking for is equal to 0.5. Can you tell why? $\endgroup$ – JohnK Jun 8 '17 at 14:02
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    $\begingroup$ @JohnK Well I see that $4.7-4.2=.5$, and $5-4 = 1$, and $\dfrac {.5}{1} = .5$ So I'm guessing this is the mgf of a uniform distribution over the interval $(4, 5)$? $\endgroup$ – Ovi Jun 8 '17 at 14:07
  • $\begingroup$ You can verify this. $\endgroup$ – JohnK Jun 8 '17 at 14:14

Generally with textbook problems like this you can figure it out by looking at a table of MGFs. Try looking here and see if that helps.

If that doesn't help, you can in principle recover the pdf from the MGF by noting the connection between MGFs and Laplace transforms. This is discussed here, for example. I can add more details if those links aren't sufficient.


From wikipedia on the Uniform distribution: The moment-generating function is:[2] $$ {\displaystyle M_{x}=E(e^{tx})={\frac {e^{tb}-e^{ta}}{t(b-a)}}\,\!} $$ In your case, b = 5 and a = 4. So the distribtuion is a uniform distribution from 4 to 5.

  • 2
    $\begingroup$ This is a self-study question, hence providing the entire solution does not help in understanding the principle behind the question. $\endgroup$ – Xi'an Jun 16 '17 at 8:51

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