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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. $\left(\dfrac{5}{3}\right)$
asked in Probability by Veteran (59.7k points)
edited by | 2k views
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 ?
+1
@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. 

0

This one ...

3 Answers

+13 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 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

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

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

$E[P^2]=E[X^2] + E[Y^2]= \frac{5}{3}$

answered by Boss (18.6k points)
selected by
+23 votes

answer = option D

answered by Boss (30.9k points)
0
Plz tell once again how u calculated the value of P(p) ?
+6

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

0
why P(p) is 1/root (5)???
+16 votes
answered by Boss (11.3k points)
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
please explain p2=x2+y2.
Answer:

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