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25 votes

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

  1. $\left(\dfrac{2}{3}\right)$
  2. $\quad 1$
  3. $\left(\dfrac{4}{3}\right)$
  4. $\left(\dfrac{5}{3}\right)$
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edited by
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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edited by

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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4 Answers

68 votes
68 votes
Best answer

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: $D$

edited by
48 votes
48 votes

answer = option D

3 Comments

reshown by
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)???
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19 votes
19 votes

3 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.
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9 votes
9 votes

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.

2 Comments

Best and most simple !
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Got JEE Vibes from this question.
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Answer:

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