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Consider two standardized random variables $x$ and $y$, and define a function

$g(x,y)=E[max(x,y)]$

where $E$ is the expected value operator.

My question is finding the upper and lower bounds of $g(x,y)$?

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  • $\begingroup$ Note that the RHS is a constant hence the LHS should NOT be written as g(x,y), which is a random variable. $\endgroup$ – Did Oct 18 '12 at 7:25
  • $\begingroup$ Something that works nicely for general x, y: observe that max(x,y)>=(x+y)/2 $\endgroup$ – Clay Thomas Oct 26 '18 at 3:00
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For nonnegative random variables $X$ and $Y$, obvious bounds are $$ \max\{\mathbb E(X),\mathbb E(Y)\}\leqslant\mathbb E(\max\{X,Y\})\leqslant\mathbb E(X)+\mathbb E(Y). $$ Without some further hypothesis, I see no reason to expect refined inequalities.

Edit: It appears that the question the OP is really interested in is:

Let $X$ and $Y$ denote two i.i.d. standard normal random variables (mean zero, variance one) and $Z=\max\{X,Y\}$, compute $\mathbb E(Z)$.

Let $\varphi(x)=\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}$ denote the standard normal PDF and $\Phi(x)=\int\limits_{-\infty}^x\varphi(t)\mathrm dt$ the standard normal CDF, then $\mathbb E(Z)=2\,\mathbb E(X\mathbf 1_{Y\leqslant X})=2\,\mathbb E(X\Phi(X))$.

Now, $x\varphi(x)=-\varphi'(x)$ and $\Phi'(x)=\varphi(x)$ hence an integration by parts leads to $$ \mathbb E(X\Phi(X))=\int_{-\infty}^{+\infty}x\varphi(x)\Phi(x)\mathrm dx=-\left[\varphi(x)\Phi(x)\right]_{-\infty}^{+\infty}+\int_{-\infty}^{+\infty}\varphi^2(x)\mathrm dx $$ Since $\varphi^2(x)=\varphi(\sqrt2x)/\sqrt{2\pi}$, the last integral is $1/(2\sqrt\pi)$ and $\mathbb E(Z)=1/\sqrt\pi$. More generally:

Let $X$ and $Y$ denote two i.i.d. centered normal random variables with variance $\sigma^2$ and $Z=\max\{X,Y\}$, then $\mathbb E(Z)=\sigma/\sqrt\pi$.

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  • $\begingroup$ the random variables $x$ and $y$ are normally distributed with mean zero, so they are not nonnegative. $\endgroup$ – Daniel Lårs Oct 18 '12 at 7:03
  • $\begingroup$ Are they independent? Which variances? $\endgroup$ – Did Oct 18 '12 at 7:04
  • $\begingroup$ they are independent $\endgroup$ – Daniel Lårs Oct 18 '12 at 7:04
  • $\begingroup$ Which variances? $\endgroup$ – Did Oct 18 '12 at 7:05
  • $\begingroup$ the variance $\sigma^2=1$ $\endgroup$ – Daniel Lårs Oct 18 '12 at 7:06

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