# GATE2004-80

3.6k views

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. $\left(\dfrac{5}{3}\right)$

edited
0
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 ?
3
@Chhotu

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

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.

1

This one ... 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 length 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 $PDF(x) = \frac{1}{1-0}=1$

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

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 0
Plz tell once again how u calculated the value of P(p) ?
9

here, $f(x) = P(p)$ 0
why P(p) is 1/root (5)???
0

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. ??

0
It is asked to calculate expectation of p^2, i.e varience of p
0

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:-

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

## Related questions

1
2.6k views
Two $n$ bit binary strings, $S_1$ and $S_2$ are chosen randomly with uniform probability. The probability that the Hamming distance between these strings (the number of bit positions where the two strings differ) is equal to $d$ is $\dfrac{^{n}C_{d}}{2^{n}}$ $\dfrac{^{n}C_{d}}{2^{d}}$ $\dfrac{d}{2^{n}}$ $\dfrac{1}{2^{d}}$
An examination paper has $150$ multiple choice questions of one mark each, with each question having four choices. Each incorrect answer fetches $-0.25$ marks. Suppose $1000$ students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is $0$ $2550$ $7525$ $9375$