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Let $𝑍=π‘‹π‘Œ$ , where $X$ and $Y$ are independent, 𝑋 ~𝐸π‘₯π‘π‘œπ‘›π‘’π‘›π‘‘π‘–π‘Žπ‘™(0.01) and π‘ŒβˆΌπ΅π‘’π‘Ÿπ‘›π‘œπ‘’π‘™π‘™π‘–(0.3)

Is there a way to find the m.g.f of 𝑍?

I know that I can find the C.D.F by doing as explained here.

Initially I tried to find de P.D.F using the C.D.F and determine the M.G.F by hand, but I can't find a way to determine the P.D.F too.

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    $\begingroup$ Since $XY$ is what is called a mixed random variable (it is neither a continuous random variable like the exponential random variable nor a discrete random variable like the Bernoulli random variable), it does not have a P.D.F. in the usual meaning of the term. The C.D.F. is discontinuous at $0$ and so you can't find the C.D.F .and then take the derivative and call the derivative the P.D.F.: the derivative is undefined at $0$. But, Yes, there is a way to find the MGF of $XY$ from first principles. $\endgroup$
    – Dilip Sarwate
    Commented Nov 27, 2021 at 21:39
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    $\begingroup$ Since $Z$ is a mixture of an Exponential distribution and an atom at $0,$ its mgf can be obtained as a mixture (with the same weights) of the mgfs of its component distributions. This follows directly from the definitions. $\endgroup$
    – whuber ♦
    Commented Nov 27, 2021 at 23:52
  • $\begingroup$ Ok, so can I solve using law of total expectation? I didn't understand what do you mean first principles. Sorry if It is a silly question, I'm relatively new to probability. $\endgroup$
    – Ismael
    Commented Nov 28, 2021 at 1:39
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    $\begingroup$ $$\mathbb E[\exp\{tXY\}]=\mathbb E_Y[\mathbb E[\exp\{tXY\}|Y]]$$ $\endgroup$
    – Xi'an
    Commented Nov 28, 2021 at 10:18
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    $\begingroup$ You start by finding noting that $XY = X$ with probability $0.3$ and $XY=0$ with probability $0.7$. Then you find the moment generating function of $X$ and the moment generating function of $0$. $\endgroup$
    – Henry
    Commented Nov 28, 2021 at 13:05

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By using law of total expectation: $$\mathbb{E}[e^{tXY}] = \mathbb{E}[\mathbb{E}[e^{tXY}|Y] = \mathbb{E}[e^{tXY}|Y=0]P(Y=0) + \mathbb{E}[e^{tXY}|Y=1]P(Y=1) = \mathbb{E}[e^{0}]0.7 + \mathbb{E}[e^{tX}]0.3 = 0.7 + 0.3M_{X}(t)$$

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    $\begingroup$ $\mbox{P}[Y=1]=.3$ not $.7$ $\endgroup$
    – user277126
    Commented Nov 30, 2021 at 4:41

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