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

$cov(x,y) = E[xy] - E[x]E[y]$

$E[xy] = E[x]E[y] + cov(x,y)$

$z' = x'y' + cov(x,y)$

covariance can be positive, negative, or zero but can we conclude anything about covariance from the given information?

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

@liontig37

The Cauchy-Schwarz inequality which you have mentioned is correct but can you please explain how did you find $E[X^2]$ and $E[Y^2]$  

Your $E[XY] \leq 1/3 $ also implies $E[XY] \leq E[X]=0.5$

If it is not justified that D is correct or there are situations where D is not correct then mention those situations.

The reason we provide proofs so that we can believe it. If there is any flaw in the proof then can you please mention it.

---

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.

---

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 ?

Answer 2

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 ?

Answer 3

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.

---

@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

---

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.