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$$ z = max(x+b,y) $$ where x ~ N(m1,s1) and y~N(m2,s2), b is a contant

What's the pdf of z?

Or exact form of E(z)? (E is expectation operator)

To the best of my guessing from the literature it is related with Weibull (https://en.wikipedia.org/wiki/Weibull_distribution) but I can't derive the exact pdf of z or exact E(z).

If there is a formula for arbitrary number of variables and covariance matrix, then that will be even better.

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Should the answer be like the following?

$$P(z) = P_2(z) \int_{-\infty}^{z-b} P_1(x)dx\; + P_1(z-b) \int_{-\infty}^z P_2(y)dy\; $$

$$ =\frac{1}{2\pi\sigma_{1}\sigma_{2}}\left[\exp\left(\frac{-(z-\mu_{2})^{2}}{2\sigma_{2}^{2}}\right)\int_{-\infty}^{z-b}\exp\left(\frac{-(x-\mu_{1})^{2}}{2\sigma_{1}^{2}}\right)dx+\exp\left(\frac{-(z-b-\mu_{1})^{2}}{2\sigma_{1}^{2}}\right)\int_{-\infty}^{z}\exp\left(\frac{-(y-\mu_{2})^{2}}{2\sigma_{2}^{2}}\right)dy\right] $$

$$ =\frac{1}{2\pi\sigma_{1}\sigma_{2}}\left[\exp\left(\frac{-(z-\mu_{2})^{2}}{2\sigma_{2}^{2}}\right)\sigma_{1}\sqrt{\frac{\pi}{2}}\mathrm{erfc}\left(\frac{\mu_{1}-z+b}{\sigma_{1}\sqrt{2}}\right)+\exp\left(\frac{-(z-b-\mu_{1})^{2}}{2\sigma_{1}^{2}}\right)\sigma_{2}\sqrt{\frac{\pi}{2}}\mathrm{erfc}\left(\frac{\mu_{2}-z}{\sigma_{2}\sqrt{2}}\right)\right] $$

Michael Medvinsky
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user1849133
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  • Something may be wrong about me, but when reading the title of this post I was like "Exact form..." (ooh! differential geometry) "...of pd..." (ooh, differential equations) "...f of" (huh? pdf? doesn't this belong in tex.se or something?) and then read the rest for an "Ooh, I see...". I should probably parse tags first. – tomasz Jul 24 '13 at 15:07

1 Answers1

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I denote the pdf of $x$ by $P_1(x)$ and the pdf of $y$ by $P_2(y)$. Then the pdf $P(z)$ of $z=\max(x+b,y)$ is given by

$$P(z) = \int_{-\infty}^{\infty} dx \int_{x+b}^\infty dy\; P_1(x) P_2(y) \delta(z-y) + \int_{y-b}^\infty dx \int_{-\infty}^\infty dy\; P_1(x) P_2(y) \delta(z-x-b)$$ $$= P_2(z) \int_{-\infty}^{z-b} P_1(x)\;dx + P_1(z-b) \int_{-\infty}^z P_2(y)\;dy$$

For normal $P_1$ and $P_2$ the integrals evaluate to error functions.

Carlo Beenakker
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  • @Carlo_Beenakker Thank you. Can you refer to the derivation of this? Some books or lecture notes or paper, whatever. And what are the deltas? What is the exact form expressions of the delta functions? – user1849133 Jul 23 '13 at 02:30
  • @Carlo_Beenakker Perhaps I am being confused of some basic notation, but.. in the 2nd line, shouldn't P1(x) and P2(y) be "in" the integral? I also guess that your remark that "For normal P1 and P2 the integrals evaluate to error functions" is clear once P1 and P2 are in the integral, not outside. – user1849133 Jul 23 '13 at 02:40
  • for the delta function see: http://en.wikipedia.org/wiki/Dirac_delta_function --- and yes, the functions of $x$ and $y$ in the second line are to be integrated. – Carlo Beenakker Jul 23 '13 at 06:39
  • @Carlo_Beenakker Thank you very much again. Maybe the derivation of this is too long to post here. So could you tell me a link to a page that I can learn from? – user1849133 Jul 24 '13 at 13:28
  • @Carlo_Beenakker And actually my question was motivated by http://pricetheory.uchicago.edu/levitt/Papers/LevittOptimalIncentiveScheme1995.pdf At page 749 the author says "The density function of the maximum of multiple draws from a normal distribution is distributed as a Type III extreme order function. For further details on the characteristics of such functions, see Galambos (1978) or Gumbel (1958)" Does your answer simplify to Type III extreme function? I tried to derive but it seemed they are very different.. – user1849133 Jul 24 '13 at 13:30
  • @Carlo_Beenakker I edited my post to plug in normal pdf. How can this be a Type III extreme function? – user1849133 Jul 24 '13 at 14:43