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Official PYQ · 2004Q80Probability: Random Variables and DistributionsMCQ1 mark

GATE CS 2004 Q80 — Probability: Random Variables and Distributions

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A point is randomly selected with uniform probability in the X−YX-Y plane within the rectangle with corners at (0,0),(1,0),(1,2)(0,0), (1,0), (1,2) and (0,2).(0,2). If pp is the length of the position vector of the point, the expected value of p2p^{2} is

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Answer: D

Let the coordinates of the randomly selected point be (X,Y)(X, Y). The point is uniformly distributed within the rectangle defined by 0≤X≤10 \le X \le 1 and 0≤Y≤20 \le Y \le 2.

Since the boundaries of XX and YY are independent, XX and YY are independent random variables with the following probability density functions (pdfs):

  • fX(x)=1f_X(x) = 1 for 0≤x≤10 \le x \le 1
  • fY(y)=12f_Y(y) = \frac{1}{2} for 0≤y≤20 \le y \le 2

The length of the position vector of the point (X,Y)(X, Y) is given by p=X2+Y2p = \sqrt{X^2 + Y^2}. Therefore, we want to find the expected value of p2p^2:

p2=X2+Y2p^2 = X^2 + Y^2

By the linearity of expectation:

E[p2]=E[X2+Y2]=E[X2]+E[Y2]E[p^2] = E[X^2 + Y^2] = E[X^2] + E[Y^2]

Now, we compute E[X2]E[X^2] and E[Y2]E[Y^2] individually:

E[X2]=∫01x2⋅fX(x) dx=∫01x2 dx=[x33]01=13E[X^2] = \int_{0}^{1} x^2 \cdot f_X(x) \, dx = \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} E[Y2]=∫02y2⋅fY(y) dy=∫02y2⋅12 dy=12[y33]02=12⋅83=43E[Y^2] = \int_{0}^{2} y^2 \cdot f_Y(y) \, dy = \int_{0}^{2} y^2 \cdot \frac{1}{2} \, dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}

Adding the two expectations together:

E[p2]=13+43=53E[p^2] = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}

Thus, the expected value of p2p^2 is 53\frac{5}{3}, which corresponds to option D.

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