# TIFR2015-A-12

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Consider two independent and identically distributed random variables $X$ and $Y$ uniformly distributed in $[0, 1]$. For $\alpha \in \left[0, 1\right]$, the probability that $\alpha$ max $(X, Y) < XY$ is

1. $1/ (2\alpha)$
2. exp $(1 - \alpha)$
3. $1 - \alpha$
4. $(1 - \alpha)^{2}$
5. $1 - \alpha^{2}$

edited

$P(\alpha \max (X, Y) < XY)$

$= P(\alpha X < XY \text{ AND } \alpha Y < XY)$

$= P(\alpha < Y \text{ AND } \alpha < X )$

$= P(\alpha < Y) \times P(\alpha < X )$ (Since, $X$ and $Y$ are independent)

$= (P(Y) - P(\alpha)) \times (P(X) - P(\alpha))$

$= (1 - \alpha) \times (1 - \alpha)$ (Uniform distribution in the interval $0$ to $1$)

$= (1 - \alpha)^2$

Option D.
0

$= (P(Y)-P(\alpha)) \times (P(X)-P(\alpha))$

$= (1-\alpha) \times (1-\alpha)$ (Uniform distribution in the interval $0$ to $1$)

@Arjun sir,

How is $P(\alpha )=\alpha$ in the above solution?

$P\left ( \alpha. max\left ( X,Y \right )< XY \right )=P\left ( \alpha X< XY \right ).P\left ( \alpha Y< XY \right )$

$=P\left ( \alpha < Y \right ).P\left ( \alpha < X \right )$

$=P\left ( Y> \alpha \right ).P\left ( X> \alpha \right )$

$=\int_{0}^{1 }\left [ 1-\alpha \right ]dx.\int_{0}^{1 }\left [ 1-\alpha \right ]dy$ [By the formula of total probability and As $X$ and $Y$ uniformly distributed between $0$ and $1$]

$=\left [ 1-\alpha \right ]^{2}$

Resources:

http://www.cs.cornell.edu/courses/cs321/2005fa/HW%20Solutions/solutions1.pdf

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/exams/MIT6_041F10_quiz02_s08_sol.pdf

https://math.stackexchange.com/questions/1484835/find-the-distribution-of-x-y-if-x-and-y-are-i-i-d-uniform-on-0-1

0
Shouldn't step 4 be $\int_{\alpha }^{1}(1-0)dx * \int_{\alpha }^{1}(1-0)dy$

Moreover, the final answer is the same.
0
@vishnu_m7 it should be

$\int_{\alpha }^{1}\frac{1}{(1-0)}dx * \int_{\alpha }^{1}\frac{1}{(1-0)}dy$

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