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My question is a follow up to this: Expected value of maximum of two random variables from uniform distribution

(Another method than the one proposed by the chosen answer is to square the c.d.f. of a single uniform (which is the c.d.f. of the maximum). Then taking the derivative to get the p.d.f. of the maximum. Then finding the expected value.)

Consider U(0,V), and lets shift down the density downwards (from the original 1/V) in the interval [0, U], and to balance it out, shift the density upwards (from the original 1) in the interval [U, V] - call this new discontinuous random variable $D$.

Now I realize two values in $D$ and I want to find the expected value of the max. However, without indicator variables (don't know how to deal with those in calculus operations), I can't come up with a closed-form c.d.f. or survival function.

Any suggestions?

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  • $\begingroup$ What you describe is NOT a discontinuous random variable. $\endgroup$
    – Did
    Oct 31, 2012 at 6:13
  • $\begingroup$ @did: Not sure how the terminology works, but i mean to say that the two segments [0, U] and [U, V] are continuous but at U there is a discontinuity/jump thats giving me headaches. $\endgroup$ Oct 31, 2012 at 6:16
  • $\begingroup$ Not sure? Why?. $\endgroup$
    – Did
    Oct 31, 2012 at 6:49

1 Answer 1

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For every i.i.d. random variables $D$, $D_1$ and $D_2$ on $(0,v)$, $$ \mathbb E(\max\{D_1,D_2\})=\int_0^v(1-\mathbb P(D\leqslant x)^2)\,\mathrm dx. $$ If the density of $D$ is $a\mathbf 1_{(0,u)}+b\mathbf 1_{'u,v)}$ for some positive $v$, $u$ in $[0,v]$ and $a=p/u$, $b=q/(v-u)$ for some $p$ in $[0,1]$ and $q=1-p$, then $$ \mathbb P(D\leqslant x)=ax\mathbf 1_{0\lt x\lt u}+(1-b(v-x))\mathbf 1_{u\lt x\lt v}, $$ hence $$ \mathbb E(\max\{D_1,D_2\})=\int_0^u(1-a^2x^2)\,\mathrm dx+\int_u^v(2b(v-x)-b^2(v-x)^2)\,\mathrm dx, $$ or, equivalently, $$ \mathbb E(\max\{D_1,D_2\})=u-\tfrac13a^2u^3+b(v-u)^2-\tfrac13b^2(v-u)^3, $$ which is also $$ \mathbb E(\max\{D_1,D_2\})=(1-\tfrac13p^2)u+(q-\tfrac13q^2)(v-u). $$

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