Give a example of discrete random variable for which mgf $M(t)=E\left(e^{tx} \right)$ does not exist . I have tried with geometric(p) distribution when $(1−p)e^t≥1$ , the mgf does not converge. Is it correct approach ?
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1You are correct that there is no convergence for that $M(t)$ when $t\ge -\log_e(1-p)$, but since that boundary is positive you can extract all the moments from the behaviour $M(t)$ around $t=0$ and so it may not be a desired response. I suspect that you are being asked either for something like the log-normal distribution (a continuous distribution) or for something where some of the moments do not exist. – Henry Jun 11 '23 at 11:11
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3You'll want a discrete distribution whose tail decreases more slowly than an exponential/geometric. – Glen_b Jun 11 '23 at 11:19
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I have found a discrete of &f(x)= (6/π^2)*(1/k^2)& where k =1,2,3,... But still have confusion about the 'k' , will it be 'x' in place of 'k' in the range ? – Oscillatory Heart Jun 11 '23 at 12:34
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1Yes, your $K$ is the equivalent of $X$, so you would want to show that $E[e^{tK}]$ or $E[e^{tX}]$ did not converge at least for positive $t$ – Henry Jun 11 '23 at 12:57
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2Welcome to CV. There is a genuine question concerning what "does not exist" means, because it is always the case that $M$ exists as a partial function: $E[e^{tX}]$ is always defined for $t=0.$ (This is discussed in the posts at https://stats.stackexchange.com/questions/389846/, for instance.) Could you please clarify? – whuber Jun 11 '23 at 15:41
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Please add the [tag: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. – kjetil b halvorsen Jun 11 '23 at 22:21