TIFR CSE 2014 | Part A | Question: 19

Question

Consider the following random function of $x$
$F(x) = 1 + Ux + Vx^{2} \bmod 5$,
where $U$ and $V$ are independent random variables uniformly distributed over $\left\{0, 1, 2, 3, 4\right\}$. Which of the following is FALSE?

• A. $F(1)$ is uniformly distributed over $\left\{0, 1, 2, 3, 4\right\}$.
• B. $F(1), F(2)$ are independent random variables and both are uniformly distributed over $\left\{0, 1, 2, 3, 4\right\}$.
• C. $F(1), F(2), F(3)$ are independent and identically distributed random variables.
• D. All of the above.
• E. None of the above.

Comments

Getting option C , as F(1) , F(2),F(3) are not independent..

 F(1) , F(2) are independent & uniformly distributed..

But F(3) depends on Combination of F(1) and F(2)..

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@Amsar How is F(3) dependent on F(1), F(2)?

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should not answer be a here

Answer 1

Here  as U and V are uniformly distributed, they will have equal probabilities for each point in {0,1,2,3,4} which by the rule of probability comes $1/5$. Now for being uniformly distributed F(1), F(2), F(3) must have same values at points {0,1,2,3,4}

So,

$F(1) = (1+U +V) mod 5 = (1 + 1/5 + 1/5) mod 5 $ for all {0,1,2,3,4}

$F(2) = (1+2U +4V) mod 5 = (1 + 2/5 + 4/5) mod 5 $ for all {0,1,2,3,4}

$F(3) = (1+3U +9V) mod 5 = (1 + 3/5 + 9/5) mod 5 $ for all {0,1,2,3,4}

so all three are uniformly distributed as they all same value for all points.

but not identical as they have values different from each other.

hence Option C is false.

Comments

When I take determinant of F1, F2 and F3 it evaluates to non zero so it is independent correct if I am wrong

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How was U and V evaluated to 1/5 ? The probability mass functions of U and V would have 1/5 values, isn't it? And the values of U and V ranging from 0 to 4.

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I also have same doubt that @Veenit has. Not getting the solution properly. 

Shivansh can you explain in a bit detail?

Answer 2

Answer will be (B)

Firstly take U=2 , V=4 , x=1

F(x)= (1+2+4) mod 5 =2

Now take U=1 , V=1 , x=2

F(x)=(1+2+4) mod 5=2

So, F(1) and F(2) not uniformly distributed over all points between {0,1,2,3,4}

Comments

F(1) and F(2) are uniformly distributed means each is uniform independently- not together.