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Let $=$ , where $X$ and $Y$ are independent, ~(0.01) and ∼(0.3)

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

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.

Ismael
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    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. – Dilip Sarwate Nov 27 '21 at 21:39
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    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. – whuber Nov 27 '21 at 23:52
  • 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. – Ismael Nov 28 '21 at 01:39
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    $$\mathbb E[\exp\{tXY\}]=\mathbb E_Y[\mathbb E[\exp\{tXY\}|Y]]$$ – Xi'an Nov 28 '21 at 10:18
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    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 $0$. – Henry Nov 28 '21 at 13:05
  • Please add the [tag:self-study] tag & read its [wiki](https://stats.stackexchange.com/tags/self-study/info). Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – kjetil b halvorsen Nov 28 '21 at 13:06
  • I think I managed to solve: $$E[e^{(tXY)}] =E[e^{(tXY)}|Y] =E[e^{(tXY)}|Y=1]P(Y=1) + E[e^{(tXY)}|Y=0]P(Y=0) =E[e^{(tX)}]P(Y=1) + E[e^0]P(Y=0) = 0.3E[e^{(tX)}] +0.7 = 0.3M_X(t) + 0.7$$. Am I doing it correctly? @kjetilbhalvorsen This is not a homework problem, I'm studying probability and started to think about the MGF of a product of an exponential and a bernoulli r.v, I've used the values just to ilustrate my question. – Ismael Nov 28 '21 at 13:30
  • Then please answer your own question, in the Answers box! – kjetil b halvorsen Nov 28 '21 at 13:31
  • "First principles" = definitions and axioms (plus elementary algebra). No theorems needed. – whuber Nov 28 '21 at 14:44

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

Ismael
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