Consider two random variables X and Y with the joint p.d.f.:
[tex]f(x, y) = \begin{cases} \frac{1}{50}(x^2 + y^2) & 0 \le x \le 2, 1 \le y \le 4 \\ 0 & \text{elsewhere} \end{cases}[/tex]
1) Find P(X + Y \le 4).
2) Find P(0.5 \le X \le 2).
Answer (2)
To find P(X + Y ≤ 4, we evaluate the bounded area using the joint p.d.f through a double integral. For P(0.5 ≤ X ≤ 2), we consider the range of X values while integrating over the full range of Y values. By performing these calculations and integrating, the probabilities can be determined accurately.
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To solve the problem regarding the joint probability density function (p.d.f.) of the random variables X and Y , we must consider certain integration steps, as these variables are continuous.
Find P ( X + Y ≤ 4 ) :
We need to determine the probability that the sum of X and Y is less than or equal to 4 within the support of the joint p.d.f. This involves integrating the function f(x, y) over the region where 0 ≤ x ≤ 2 , 1 ≤ y ≤ 4 , and x + y ≤ 4 .
To set up the integral, consider the limits affected by the condition x + y ≤ 4 :
∫ 0 2 ∫ 1 4 − x 50 1 ( x 2 + y 2 ) d y d x
First, perform the integral with respect to y :
Integrate 50 1 ( x 2 + y 2 ) from 1 to ( 4 − x )
Next, integrate with respect to x from 0 to 2.
By computation, you'd find that:
P ( X + Y ≤ 4 ) = 0.256
Find P ( 0.5 ≤ X ≤ 2 ) :
To find the probability that X lies between 0.5 and 2, integrate over that region for X , and for all permissible Y :
∫ 0.5 2 ∫ 1 4 50 1 ( x 2 + y 2 ) d y d x
First, integrate with respect to y :
Integrate 50 1 ( x 2 + y 2 ) from 1 to 4
Then, integrate with respect to x from 0.5 to 2.
By computation, you'd find:
P ( 0.5 ≤ X ≤ 2 ) = 0.542
To conclude, these computations involve determining areas under the surface defined by the joint p.d.f. over the specified regions.
