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Let $X$ be a random variable distributed uniformly over the interval $[0,2]$ and let $Y$ be a random variable distributed exponentially with parameter $1$. Describe a probability space in which $$P(X+Y>2)\geq0.5$$

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Assume that $X$ is uniformly distributed on $(0,2)$ and define $Y$ as a function of $X$ as follows:

  • If $X\lt1$, then $\mathrm e^{-Y}=\frac12X+\frac12$.
  • If $X\gt1$, then $\mathrm e^{-Y}=\frac12X-\frac12$.

Thus, $\mathrm e^{-Y}$ is uniform on $(\frac12,1)$ in the first case and uniform on $(0,\frac12)$ in the second case. Since $X$ is uniform on $(0,2)$, the two cases are equiprobable, hence $\mathrm e^{-Y}$ is (globally) uniform on $(0,1)$ or, equivalently, $Y$ is exponentially distributed with parameter $1$. Now, it happens that $$ [X+Y\gt2]=[X\gt1]=[X+Y\gt2+\log2], $$ hence there exist some random variables $X$ and $Y$ with the desired distributions such that $\mathbb P(X+Y\geqslant2+\log2)=\frac12$, with $2+\log2=2.69^+$.

To show the claimed property, assume that $X\gt1$, then $X+Y=u(Y)$ with $\mathrm e^{-Y}$ in $(0,\frac12)$ and $u:y\mapsto2\mathrm e^{-y}+1+y$. In particular, $Y\gt\log2$ and the function $u$ is increasing on the interval $(\log2,+\infty)$ hence $u(Y)\gt u(\log2)=2+\log2$ on $[X\gt1]$.

This shows that $[X\gt1]\subseteq[X+Y\gt2+\log2]$, which is enough for our purpose (even though the claimed identity $[X\gt1]=[X+Y\gt2+\log2]$ holds).

Edit: Here is a variant of this construction. Let $a$ in $(0,\frac12)$. Assume once again that $X$ is uniformly distributed on $(0,2)$ and define $Y$ as a function of $X$ as follows:

  • If $X\lt 2a$, then $\mathrm e^{-Y}=\frac12X+1-a$.
  • If $X\gt 2a$, then $\mathrm e^{-Y}=\frac12X-a$.

If $X\gt 2a$, then $X+Y=u_a(Y)$ with $\mathrm e^{-Y}$ in $(0,1-a)$ and $u_a:y\mapsto2\mathrm e^{-y}+2a+y$. In particular, $Y\gt y_a$ with $y_a=-\log(1-a)$ and the function $u_a$ is decreasing on the interval $(y_a,\log2)$ and increasing on the interval $(\log2,+\infty)$, with $u_a(\log2)=1+\log2+2a$. Thus, $u_a(Y)\gt 1+\log2+2a$ on $[X\gt2a]$, and $$ \mathbb P(X+Y\gt 1+\log2+2a)\geqslant 1-a. $$ Thus, every $a$ in $(\frac12(1-\log2),\frac12)$ achieves $\mathbb P(X+Y\geqslant x)\geqslant p$ for some $x\gt2$ and $p\gt0.5$ (our first example being the case $a=\frac12$, for which $x=2+\log2=2.69^+$ and $p=0.5$). For example, $a=\frac12-\frac14\log2$ yields numerically $$ \mathbb P(X+Y\gt\tfrac73)\gt\mathbb P(X+Y\gt2.34)\gt0.67\gt\tfrac23. $$

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+1 Neat! $\phantom{neat indeed}$ –  Sasha Oct 29 '12 at 14:50
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This is funny, my box of comments indicates four comments on this answer, but only one seems to have survived... –  Did Oct 29 '12 at 19:37
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@did I had to retract one too. I was trying to suggest a way for the probability to surpass the value of $1/2$, but then realized by my construction does not have correct marginals. –  Sasha Oct 29 '12 at 20:39
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@Sasha Your comment made me look at the matter again, see Edit (and thanks). –  Did Oct 29 '12 at 21:19
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@EvanAad the strategy appears to be that we take $\Omega = [0, 2]$ and $X(\omega) = \omega$; then, for $Y$, we alter slightly the fact that $-\log (1 - \frac X 2) \sim Exp(1)$; starting at $\omega \ge 1$ we glue the part of the exponential near infinty to $\omega = 1$ and the part near $\log(2)$ to $2$; this leads to $Y(\omega) = -\log\frac{\omega - 1}{2}$ for $\omega > 1$ instead of taking $-\log(1 - \frac \omega 2)$ on the whole interval. We are shuffling around the tails of the exponential to move the big part to where we need it ($\omega = 1$) and moving less to $\omega = 2$ where e don't. –  guy Oct 29 '12 at 22:35
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