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The joint probability density function of X and Y is given as f(x,y)=e^-{x+y}


We give the solution to the following problem. Suppose that the joint probability density function X and Y is given as f(x,y)=e^-{x+y}. Then find:

  1. P(X=Y)
  2. P(min(X,Y)>1/2)
  3. P(X ≤ Y)
  4. the marginal density function of Y.
  5. E[XY]


We are given that X and Y have the following joint PDF,

f(x,y)=e^-{x+y} Step 1

1) We define a new variable Z = X – Y. Since X and Y are random variables so is Z. Now, P(X=Y) = P(X-Y=0) = P(Z=0).

But we know that the probability of a random variable taking a simple point value is always zero. Therefore, P(Z=0) = 0 which is the required answer.

2) The minimum value of X and Y is greater than 1/2 if and only if both X and Y themselves take values greater than 1/2.

This is because if either X or Y were less than 1/2 then their minimum value would definitely be less than 1/2. Therefore we have,

f(x,y)=e^-{x+y} Step 2

3) In order to calculate P(X ≤ Y), we let X vary between 0 and Y, whereas Y varies freely between 0 to infinity.

Therefore we have,

f(x,y)=e^-{x+y} Step 3

4) In order to find the marginal probability density function of Y, we integrate the joint PDF over the random variable X.

f(x,y)=e^-{x+y} Step 4

So the marginal density function of Y is,

f(x,y)=e^-{x+y} Step 5

5) We use the formula,

f(x,y)=e^-{x+y} Step 6

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