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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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4 Answers

Best answer
78 votes
78 votes

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: $D$

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answer = option D

19 votes
19 votes
12 votes
12 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.

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