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I need some help people. I have a logistic r.v. $V\sim \Lambda (0, \frac {\pi^2}{3})$ and a Bernoulli $Z \sim B(p_z)$, independent. I have also $c\in R$, $\tilde Z = cZ$, and I have to find the density of $$U = V + \tilde Z$$

I know that one can use the convolution method by using the dirac delta function to write the pmf of $\tilde Z$ as a weighted sum, but I have trouble with the nuts and bolts, I couldn't find something clear in the books, while some related posts I found here and in math.SE didn't do the trick for me.

$1)$ Is $$P(\tilde Z=\tilde z) = p_z\delta (\tilde z+c) + (1-p_z)\delta(\tilde z)$$ the correct way to write the pmf of $\tilde Z$?

and,

$2)$ in the convolution integral should I make the substitution $u =v+\tilde z \Rightarrow v=u- \tilde z$ and integrate w.r.t to $\tilde z$, or $u =v+\tilde z \Rightarrow \tilde z=u-v$ and integrate w.r.t to $v$?

I have a feeling this is trivial, but this time I did go blind.

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    $\begingroup$ It's just a straight finite mixture of V with prob 1-p and V shifted up by c, with prob p. both the pdf and the cdf are straightforward linear combinations of the two components. (Don't sweat it, you're probably just having one of those days where things don't seem to go smooth. For me that's most days) $\endgroup$
    – Glen_b
    Commented Nov 17, 2013 at 1:06
  • $\begingroup$ Thanks @Glen_b. I' ll let it cool-off. Do you mean by any chance "shifted by $c$" , instead of shifted by $1$, since the original Bernoulli is multiplied by $c$? $\endgroup$
    – Alecos Papadopoulos
    Commented Nov 17, 2013 at 1:10
  • $\begingroup$ I did. I caught it and fixed it before you finished typing, even, and added a link. Actually, I'd have posted it as a short answer, but once you're sorted out, you'll be able to write a much better answer than I would; your mathematical statistics generally looks to be better than mine. $\endgroup$
    – Glen_b
    Commented Nov 17, 2013 at 1:10
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    $\begingroup$ @Glen_b Thanks! - and thanks also for the link. And I commit to post the answer. $\endgroup$
    – Alecos Papadopoulos
    Commented Nov 17, 2013 at 1:13

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(As promised, I am posting the answer to my own question. And there is no need for... convoluted mathematical statistics here, it is indeed straightforward, but it only was revealed to be so because of @Glen_b comments).

The cdf of $U$ can be written $$F_U(u) = P(U \le u) = P(U \le u\mid \tilde Z =c)\cdot P(\tilde Z =c) + P(U \le u\mid \tilde Z =0)\cdot P(\tilde Z =0)$$

$$=p_zP(U \le u\mid \tilde Z =c) + (1-p_z)P(U \le u\mid \tilde Z =0)$$

Now when $\tilde Z =c \Rightarrow U=V+c$ while when $\tilde Z =0 \Rightarrow U=V$. So

$$F_U(u) = p_zP(V+c \le u) + (1-p_z)P(V \le u)$$

Since $V\sim \Lambda (0, \frac {\pi^2}{3})\equiv \Lambda \Rightarrow V+c\sim \Lambda (c, \frac {\pi^2}{3})\equiv\Lambda_{c} $. Therefore

$$F_U(u) = p_zΛ_{c}(u) + (1-p_z)\Lambda(u)\\=p_z\Big(1+\exp\{-(u-c)\}\Big)^{-1}+(1-p_z)\Big(1+\exp\{-u\}\Big)^{-1}$$

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  • $\begingroup$ +1 Looks like I missed the original posting of the answer. $\endgroup$
    – Glen_b
    Commented Oct 7, 2014 at 3:04

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