GATE CSE 2024 | Set 2 | Question: 34

Question

Let $x$ and $y$ be random variables, not necessarily independent, that take real values in the interval $[0,1]$. Let $z=x y$ and let the mean values of $x, y, z$ be $\bar{x}, \bar{y}, \bar{z}$, respectively. Which one of the following statements is TRUE?

• A. $\bar{z}=\bar{x} \bar{y}$
• B. $\bar{z} \leq \bar{x} \bar{y}$
• C. $\bar{z} \geq \bar{x} \bar{y}$
• D. $\bar{z} \leq \bar{x}$

Comments

Ans

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If X ≤ Y for all possible outcomes, E[X] ≤ E[Y] → Monotonicity

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x and y bounds in [0,1] so,

If y∈[0,1], then 0 ≤ xy ≤ x because x≥0

Similarly, 0 ≤ xy ≤ y

Answer 1

Answer Should be Option (D).

Lets first give counter example for (C) and (A):

Lets assume $X$ is a random variable uniformly distributed in $[0,1]$ and $Y$ is $1-X$ .

So, $E(X)=\frac{1}{2}$

and $E(Y)=E(1-X)=E(1)-E(X)=1-\frac{1}{2}=\frac{1}{2}$

Now,Given $Z=XY$

$E(Z)=E(XY)= \int_{0}^{1} x(1-x) \,dx=\frac{1}{6}$ 

So, $E(Z) \leq E(X)E(Y)$ 

[Note here Covariance is negative]

Counter Example for (B):

Lets assume $X$  is a random variable uniformly distributed in $[0,1]$ and $Y=X$.

So, $E(X)=\frac{1}{2}$

$E(Y)=\frac{1}{2}$

Now Given $Z=XY$

So,$E(Z)=E(XY)=\int_{0}^{1} x*x \,dx=\frac{1}{3}$

So, $E(Z) \geq E(X)E(Y)$ 

[Note here Covariance is positive] 

Now , why does option (D) is correct ?

=> It is given , $X$ and $Y$ takes value in the range $[0,1]$.

So, we can write , $0 \leq Y \leq 1$ .

Multiply this inequality by $X$.

So, $0 \leq XY \leq X$ 

Now take expectation of this Inequality,

So, $E(0) \leq E(XY) \leq E(X)$ 

So, $0 \leq E(XY) \leq E(X)$

Which proof the option (D) .

Note this inequality Holds true because $X$ and $Y$ takes value in the range between $[0,1]$.

Comments

how did i find E(X^2) ? I mentioned unfirmly distributed in the interval [0,1].

So <=0.3 justifies <=0.5 ? How is value 0.4 valid ?

For example kindly see comment section below too. One person has already mentioned the same.

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How do you find $E[xy]=0.4  \ ?$

Definition says  $E[XY]=\int_{-\infty}^{\infty}xy \ f_{X,Y}(x,y) \ dx \ dy$

To find $E[XY],$ you need the joint pdf of X and Y.

So, what is your joint pdf $f_{X,Y}(x,y)$ to get $E[xy]=0.4 \ ?$

I have provided one example above in which $E[x]=\frac{7}{12}$ but here also $x$ is uniformly distributed in $[0,1]$ so why I have not written 0.5 ?

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Instead of assuming the distributions we can think of the ranges in the question.

We know that x is between 0 and 1 and so is Y. In that case XY would naturally be a number that is lesser than or at most equal to X or Y. This is only for the range [0,1] which means that the expectations of these variables would also follow suite

hence E(XY) <= E(X)

Which makes sense on an average xy would be less than x only an extreme case will make xy = x.

Please do correct me if I am thinking wrong.

Answer 2

Ans

Answer 3

Explanation :

As given in the question, X and Y can take values in the interval of [0,1]. From this, we can write 0 <= Y <= 1. 

This can also be rewritten as 0 <= XY <= X.

E(X–c) = E(X)–c where c is a constant. Also called: mathematical expectation.

Taking the expectation of the above inequality, we will get this:

E(0) <= E(XY) <= E(X)

This is also equal to 0 <= E(XY) <= E(X)

This can only be true in the case when the range is specified between 0 and 1.

Therefore, the correct option is D.

Answer 4

E(Z) = E(X)E(Y) + cov(X,Y)

if E(X)=1/2, E(Y)=1, cov(X,Y)= 1/4

option D will be false.

option A can’t be true since variables are not necessarily independent.

Out of options B and C cov(X,Y) depends on relation between X and Y. If X and Y are proportional covariance will be positive else it is negative.

It is given that Z = XY or Y = Z/X. For a particular value of Z if X increases then Y decreases. This makes relation between X and Y as inversely proportional. So cov(X,Y) is negative.

This makes option B true.

Comments

When we talk about correlation or covariance between X and Y, why does Z come into the picture?

Z = XY has NOTHING to do with the correlation between X and Y.

Imagine along with Z, I define T = X/Y, now does it change anything about the correlation between X and Y ?

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Z=XY has to do with correlation/covariance between X and Y but not T=X/Y.

Answer 5

Naive approach to get answer : 

Answer 6

Options A , B, and C are not always true. But I want to show in answer that Answer D also can be wrong in some cases and not always true.

So option D can also be wrong.

Comments

@Arjun Sir, @deepakpoonia sir, @Sachinmittal1 sir, please explain if my analogy is incorrect.

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@Argharupa Adhikary Your analogy is correct. Option D can also be wrong. 

Both options B and D should be given marks. This question needs to be challenged.

@Deepak Poonia Sir, @Sachin Mittal 1 Sir have a look

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if x is distributed from 0 to 1 and y = 2x, then y is distributed from 0 to 2 and not 0 to 1 which contradicts the given information that x and y take values from 0 to 1.