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The two-dimensional distribution of the random variables $\xi$ and $\eta$ is specified by the distribution function $$ \mathbb{P}(\xi \leq x,\; \eta \leq y) = F(x,y) = \begin{cases} 0 & \textrm{ if } \min(x,y) <0 \\ \min(x,y) & \textrm{ if } 0 \leq \min(x,y) \leq 1 \\ 1 & \textrm{ if } \min(x,y) > 1 \end{cases}.$$ Find $\mathbb{P}[ (\xi-\frac{1}{2})² + (\eta -\frac{1}{2})² \leq \frac{1}{4}]$.

My target is to find the required probability. But I am stuck in the middle way.

In order to find the probability this I did $$ 0≤min(x,y) ≤1$$
$$0≤F(x,y)≤1$$
Let us denote the events $$ξ≤x$$ as $$A$$ and $$η≤y$$ as $$B$$
Now $AB⊆ A$ and $AB⊆ B$
Therefore $P(AB)≤P(A) , P(AB)≤P(B)$
Therefore $[P(AB)]² ≤P(A). P(B)$
Therefore $[P(ξ≤x, η≤y)]² ≤ P(ξ≤x). P(η≤y)$
Therefore $[F(x,y)]² ≤ F(x). F(y)$

Now if $ξ$ has the distribution function $F(x)$ and $η$ has the
distribution function $F(y)$ and if they are independent then I
can proceed to find the required probability.

By the way these $F(x), F(y)~Uniform(0,1)$ with mean $1/2$
Then $T = ξ-1/2$ ~ $N(0,1/12)$
and $W = η-1/2$ ~ $N(0,1/12)$

Then $P[12(T²+W²) ≤3]$ where $12(T²+W²)$ ~ Chi-Square Distribution with d.f 2

Is it the correct way?

If the independence does not work should I use this formula
$F(x,y)+F(0,0)-F(x,0)-F(0,y)$
Please Help Me.

vbm
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  • I may be mistaken, but for $F(x,y)$ to be a probability distribution function doesn't it need the property that $\int_{-\infty}^{\infty}\int_{-\infty}^\infty F(x,y) dydx = 1$? It would seem to me that for your $F(x,y)$ you would have this diverge. – Benji Altman Dec 14 '17 at 21:19
  • @BenjiAltman No, not the distribution function, but the probability density function must satisfy that. – Shashi Dec 14 '17 at 21:25
  • @Shashi Alright, I'm going to have to review my probability textbook then, thanks. – Benji Altman Dec 14 '17 at 23:12

1 Answers1

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  • The marginal distributions of $\xi$ and $\eta$ are indeed $\text{Unif}(0,1)$. But they are not independent: if they were, then $F(x,y) = xy$ for $x,y \in [0,1]$.
  • Regarding your attempts to show independence: In general, independence of $\xi$ and $\eta$ is equivalent to $F_{\xi,\eta}(x,y) = F_\xi(x) F_\eta(y)$.
  • I have no idea where you got the chi-squared distribution to appear.

Hint: the biggest step you can take here is to identify in plain words what the joint distribution of $(\xi,\eta)$ is.

Big, big hint:

Your CDF is the CDF of the following distribution: $\xi \sim U(0,1)$ and $\eta = \xi$. Note that this does not mean "$\eta$ is another $U(0,1)$ random variable independent of $\xi$", but rather "$\mathbb{P}(\xi=\eta)=1$."

Check that you understand how to verify the above hint. From there, you should be able to complete the computation easily.

Thus, $\mathbb{P}\{(\xi-\frac{1}{2})^2 + (\eta - \frac{1}{2})^2 \le \frac{1}{4}\} = \mathbb{P}\{(\xi - \frac{1}{2})^2 \le \frac{1}{8}\} = \mathbb{P}\{\frac{1}{2} - \frac{1}{\sqrt{8}} \le \xi \le \frac{1}{2} + \frac{1}{\sqrt{8}}\} = \frac{1}{\sqrt{2}}$.

angryavian
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  • The solution you did I understood except one little issue. One thing to confirm that since marginal distribution of $ξ$ and $η$ are same $U(0,1)$, is that why $P(ξ=η)=1$? .. Help me to clear this particular one. – vbm Dec 14 '17 at 22:22
  • @thevbmthelord No, the are many other joint distributions where the marginal distributions are each $U(0,1)$, so that is not enough. In order to prove the hint, you need to use the specific definition of $F(x,y)$ given in your question. – angryavian Dec 14 '17 at 22:29