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

@Kabir5454 can you tell why you didn't use double integration for calculating $E[XY]$

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Still answer will be same i think because in that case for $y$ limit will be $0$ to $x$ and for $X$ limit will be 0 to 1 as X and Y is dependent.

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Sir, it can happen that mean of X is 1/4, mean of Y is 1 and we know E(XY) = E(X) * E(Y) + Cov(XY). Cov(XY) can be positive negative zero anything. So E(XY) can be greater than E(X) also. So there is possibility that option D also wrong.

I have learnt from many sites that, if X and Y not necessarily independent then if X <= Y then E(X) <= E(Y) not always true.

Is my reasoning correct please explain. @deepakpoonia @sachinmittal1 @Arjun sir

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@Argharupa Adhikary your counter-example for option D is not mathematically rigid...make a strong counter-example like @Kabir5454 did for options A, B, and C..although I'm not sure if he calculated E[XY] correctly... @ankitgupta.1729 sir, can you please verify the answer..

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Tell me one thing as per cauchy schwartz inequality:

E(XY)<= sqrt(E(X^2)E(Y^2))

E(X)= 0.5,E(Y)=0.5

E(X^2)= 1/3 E(Y^2) = 1/3

So their product is 1/9

E(XY)<= sqrt(1/9)

E(XY)<= 1/3

Now how is option D which is E(X’)= 0.5 valid here ?

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@Suraj Reddy

D is correct as it is proved.

Counter-example for A and B:

https://gateoverflow.in/?qa=blob&qa_blobid=12102003270788778847

https://gateoverflow.in/?qa=blob&qa_blobid=17818555496834266689

[Edit:] The joint pdf can be taken as $f(x,y)=8xy$ instead of $g(x,y)=xy$ for the given interval because $f(x,y)=8xy$ is not a valid joint pdf. 

Conclusion remains the same but in the calculation of $E[x],E[y]$ and $E[xy]$ multiplication of $8$ should be there  and hence $E[xy]=8/18,$ $E[y]=8/10$ and $E[x]=8/15$

credit @ https://gateoverflow.in/423666/gate-da-2024-answer-key-challenge

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@ankitgupta.1729 bhaiya is there any calculation mistake in my calculation ? is my counter example for A and B wrong ?

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@Kabir5454 

You have not defined the joint pdf for the random vector $(x,y),$ if it exists.

Like with the single random variable say $X$ and $Y=g(X)$ for some function $g,$ and we define the interval or the set of values for $x$ as well as the pdf/pmf and

then we write $E[Y]=\Sigma_{x \in S_X}g(x)p_X(x)$ or $E[Y]=\int_{-\infty}^{\infty}g(x)f_X(x) dx$ when the expectation exist i.e. $\Sigma_{x \in S_X}|g(x)|p_X(x) <\infty$ or $\int_{-\infty}^{\infty}|g(x)|f_X(x)  \ dx <\infty.$

It is called the absolute convergence of the summation. ($S_X$ is the support of $X$ which represents the points of X which have the positive probability)

So, It is also possible that expectation of a random variable does not exist. For example, Cauchy Distribution.

Same goes with more than one random variable.

Suppose, $(X_1,X_2)$  is a random vector of the continuous type and $Y=g(X_1,X_2)$ is a real-valued function then

$E[Y]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x_1,x_2)f_{X_1,X_2}(x_1,x_2) \ dx_1 \ dx_2$

If $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}|g(x_1,x_2)|f_{X_1,X_2}(x_1,x_2) \ dx_1 \ dx_2 < \infty$

Here, $f_{X_1,X_2}(x_1,x_2)$ is called the Joint Probability Density Function (pdf) with the given interval for $x_1$ and $x_2$ and from this we define the Marginal pdf $f_{X_1}(x_1)$ and $f_{X_2}(x_2)$.

So, when we talk about the random vector $(X_1,X_2)$ we talk about all these things.

You have defined $Y=1-X$ which is called a transformed random variable which has same cdf as $X$ and so they are called "equal in distribution" though $X \neq Y.$ 

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You have already proved the given claim very nicely and probably someone could see this also.  

As counter-example for C is not given, So I am giving it but probably I should not have to give it because you have already proved the claim.

Counter-example for C:

$f(x,y)=x+y$ for $0 \leq x \leq 1$ and $0 \leq y \leq 1$

and $f(x,y)=0$ otherwise

Here, $E[x]=E[y]=\frac{7}{12}$ and $E[xy]=\frac{1}{3}$

(Sorry for the long comment)    

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@ankitgupta.1729  OP bhaiya 

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There are situations even where option D is not true. One such example is mentioned below in comments. Also is cauchy schwartz inequality incorrect ? Can E(XY) = 0.45 ? Just because you or some people marked option D doesn’t mean it is justified in every aspect.

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@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.

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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 ?

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.

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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.