Timeline for MGF of the product of a exponential and a bernoulli random variable
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2022 at 16:46 | vote | accept | Ismael | ||
| Nov 29, 2021 at 4:01 | history | edited | Ismael |
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| Nov 29, 2021 at 0:06 | answer | added | Ismael | timeline score: 1 | |
| Nov 28, 2021 at 14:44 | comment | added | whuber♦ | "First principles" = definitions and axioms (plus elementary algebra). No theorems needed. | |
| Nov 28, 2021 at 13:31 | comment | added | kjetil b halvorsen♦ | Then please answer your own question, in the Answers box! | |
| Nov 28, 2021 at 13:30 | comment | added | Ismael | 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. | |
| Nov 28, 2021 at 13:06 | comment | added | kjetil b halvorsen♦ | Please add the self-study tag & read its wiki. 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. | |
| Nov 28, 2021 at 13:05 | comment | added | Henry | 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 $0$. | |
| Nov 28, 2021 at 13:03 | history | edited | kjetil b halvorsen♦ |
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| Nov 28, 2021 at 10:32 | review | Close votes | |||
| Nov 28, 2021 at 17:43 | |||||
| Nov 28, 2021 at 10:18 | comment | added | Xi'an | $$\mathbb E[\exp\{tXY\}]=\mathbb E_Y[\mathbb E[\exp\{tXY\}|Y]]$$ | |
| Nov 28, 2021 at 1:39 | comment | added | Ismael | 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. | |
| Nov 27, 2021 at 23:52 | comment | added | whuber♦ | 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. | |
| Nov 27, 2021 at 21:39 | comment | added | Dilip Sarwate | 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. | |
| S Nov 27, 2021 at 20:49 | review | First questions | |||
| Nov 27, 2021 at 21:38 | |||||
| S Nov 27, 2021 at 20:49 | history | asked | Ismael | CC BY-SA 4.0 |