GATE CSE 2004 | Question: 80

Question

A point is randomly selected with uniform probability in the $X-Y$ plane within the rectangle with corners at $(0,0), (1,0), (1,2)$ and $(0,2).$ If $p$ is the length of the position vector of the point, the expected value of $p^{2}$ is

• A. $\left(\dfrac{2}{3}\right)$
• B. $\quad 1$
• C. $\left(\dfrac{4}{3}\right)$
• D. $\left(\dfrac{5}{3}\right)$

Comments

I have one doubt in the posted solutions. We have one more ways to calculate probability of each point in 2-D plane  $\rightarrow$
For ex. P = $\frac{1}{2}$  will come via two points $\left ( \frac{1}{2}, 0 \right ) and \left (0, \frac{1}{2} \right )$ respectively. Now probability of obtaining each point is $\frac{1}{2}*1$ and $\frac{1}{2}*1$ (because X and Y both are following uniform distribution and both are independent). So total probability of getting P =  $\frac{1}{2}$ is $\left ( \frac{1}{2} + \frac{1}{2} \right )$ . But via mentioned method probability for P =  $\frac{1}{2}$ will be $\frac{1}{\sqrt{5}}$ .

Where am i doing mistake ?

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@Chhotu

First of all considering uniform random variable which is continuous the probability at a point is 0.So, you cannot solve like this.

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Thank you @VS ji,

But first of all my following assumption is wrong ->

For ex. P = 1/2  will come via two points (1/2,0)and(0,1/2) respectively.

P = 1/2 will come via all the points in quarter circle whose radius is 1/2. But if we will do via this approach then we will not be able to cover all points in mentioned rectangle. 

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Ans

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Crux of problem : 

1) X and Y are independent random variables here .

X is uniformly distributed on [0,1].

Y is uniformly distributed on [0,2].

2) Apply the concept of Linearity of Expectation on P^2 = X^2 + y^2 and then calculate each expectation.

 

Answer 1

Above diagram depicts the scenario of our Question.

The length $p$ of our point $(x,y)$ selected randomly in $XY$ plane, from origin is given by

$p=\sqrt{x^{2} + y^{2}}$

$p^{2} = x^{2} + y^{2}$

Expected value of $p^{2}$ is given by

$E[p^{2}]=E[x^{2} + y^{2}]$

By linearity of expectation

$E[x^{2} + y^{2}]=E[x^{2}]+E[ y^{2}]$

Now we need to calculate the probability density function of $X$ and $Y.$

Since, distribution is Uniform

$X$ goes from $0$ to $1,$ so $\textsf{PDF}(x) = \frac{1}{1-0}=1$

$Y$ goes from $0$ to $2$ so $\textsf{PDF}(y) = \frac{1}{2-0}=\frac{1}{2}$

Now we evaluate

$E[X^2]= \int_{0}^{1}x^{2}.1dx = \frac{1}{3}$

$E[Y^2]=\int_{0}^{2}y^{2}\times(1/2)dy = \frac{4}{3}$

$E[P^2]=E[X^2] + E[Y^2]= \frac{5}{3}$

Correct Answer: $D$

Comments

YES!!

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uss bro

 

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bruhh

 

Answer 2

answer = option D

Comments

Plz tell once again how u calculated the value of P(p) ?

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here, $f(x) = P(p)$

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why P(p) is 1/root (5)???

Answer 3

Let A = (x,y) be any point in that rectangular plane. The length of the position vector p is $p = \sqrt{x^{2} + y^{2}}$

Therefore,

$E(p^{2}) = E(x^{2}+y^{2}) = E(x^{2})+E(y^{2})$              ...........(1)

Now, x and y are random variables distributed uniformly over (0,1) and (0,2) respectively.

We know, that for a random variable X distributed uniformly over (a,b),

$E(X) = \frac{a+b}{2}$ and $Var(X) = \frac{(b-a)^{2}}{12}$

Also, for any random variable, in general, $Var(X) = E(X^{2}) - E(X)^{2}$

For the random variable x,

$E(x) = \frac{0+1}{2} = \frac{1}{2}$

$Var(x) = \frac{(1-0)^{2}}{12} = \frac{1}{12}$

$\therefore E(x^{2}) = Var(x) + E(x)^{2} = \frac{1}{12}+\frac{1}{4} = \frac{1}{3}$              ...........(2).

 

For the random variable y,

$E(y) = \frac{0+2}{2} = 1$

$Var(y) = \frac{(2-0)^{2}}{12} = \frac{1}{3}$

$\therefore E(y^{2}) = Var(x) + E(x)^{2} = \frac{1}{3}+1 = \frac{4}{3}$              ...........(3)

Using (2) and (3) in (1), we get:-

$E(p^{2}) = E(x^{2}) + E(y^{2}) = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$

Hence, the answer is D.

Comments

Best and most simple !

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Got JEE Vibes from this question.

Answer 4

See Q.80 here : http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2004.html

Comments

2 doubt-

1. why we are taking P(p)dp while calculating E(p^2), why not P(p^2)dp  ?? 

2. why can't we first calculate expected value of P, and then take square of it. ??

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It is asked to calculate expectation of p^2, i.e varience of p

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please explain p2=x2+y2.

Answer 5

Ans

Answer 6

The correct answer is D. (5/3).

 

Step-by-Step Solution

 

1. Define the Variables and Goal

   • A point $(X, Y)$ is chosen uniformly from the rectangle defined by $0 \le x \le 1$ and $0 \le y \le 2$.

   • This means $X$ and $Y$ are independent random variables.

     • $X$ is uniformly distributed on $[0, 1]$.

     • $Y$ is uniformly distributed on $[0, 2]$.

   • $p$ is the length of the position vector, so $p = \sqrt{X^2 + Y^2}$.

   • We need to find the expected value of $p^2$.

      

     $$p^2 = \left(\sqrt{X^2 + Y^2}\right)^2 = X^2 + Y^2$$

2. Use Linearity of Expectation

   The goal is to find $E[p^2] = E[X^2 + Y^2]$.

   A key property of expectation is $E[A + B] = E[A] + E[B]$. Therefore:

    

   $$E[X^2 + Y^2] = E[X^2] + E[Y^2]$$

    

   We can solve for $E[X^2]$ and $E[Y^2]$ separately.

3. Calculate $E[X^2]$

   For a continuous variable $X$ with probability density function (PDF) $f(x)$, $E[X^2] = \int x^2 f(x) \,dx$.

   • The PDF for $X$ (uniform on $[0, 1]$) is $f(x) = \frac{1}{1-0} = 1$.

   • $E[X^2] = \int_{0}^{1} x^2 \cdot 1 \,dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

4. Calculate $E[Y^2]$

   • The PDF for $Y$ (uniform on $[0, 2]$) is $f(y) = \frac{1}{2-0} = \frac{1}{2}$.

   • $E[Y^2] = \int_{0}^{2} y^2 \cdot \frac{1}{2} \,dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_{0}^{2} = \frac{1}{2} \left( \frac{2^3}{3} - \frac{0^3}{3} \right) = \frac{1}{2} \left( \frac{8}{3} \right) = \frac{8}{6} = \frac{4}{3}$

5. Sum the Results

   Now we add the two expected values together:

    

   $$E[p^2] = E[X^2] + E[Y^2] = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$$