Let the joint density function of the random variables X and Y be
f(X,Y)(x,y)= 2xe^(x(^2)-y) for 0<x<1 , y>x^2
0 otherwise
(a) Find the marginal density function of X.
(b) Find the marginal density function of Y .
(c) Compute the probability P(Y < 3X^2)
You might start here. Where do you get stuck?
https://math.stackexchange.com/questions/248751/how-do-i-find-the-marginal-probability-density-function-of-2-continuous-random-v
To find the marginal density functions of X and Y, we need to integrate the joint density function f(X,Y)(x,y) over the entire range of the other variable.
(a) Marginal density function of X:
To find the marginal density function of X, we integrate the joint density function over the range of Y.
∫[x^2,∞] 2xe^(x^2 - y) dy
To evaluate this integral, we need to split it into two parts:
∫[x^2,∞] 2xe^(x^2 - y) dy = ∫[x^2,∞] 2xe^(x^2) e^(-y) dy
= 2xe^(x^2) ∫[x^2,∞] e^(-y) dy
Integrating e^(-y) from y = x^2 to y = ∞ gives:
= 2xe^(x^2) * [-e^(-y)] evaluated from x^2 to ∞
= 2xe^(x^2) * (0 - (-e^(-x^2)))
= 2xe^(x^2) * e^(-x^2)
= 2x
Therefore, the marginal density function of X is:
f_X(x) = 2x for 0 < x < 1
= 0 otherwise
(b) Marginal density function of Y:
To find the marginal density function of Y, we integrate the joint density function over the range of X.
∫[0,1] 2xe^(x^2 - y) dx
To evaluate this integral, we can treat y as a constant:
∫[0,1] 2xe^(x^2 - y) dx = 2e^(-y) ∫[0,1] xe^(x^2) dx
Integrating xe^(x^2) from x = 0 to x = 1 gives:
= 2e^(-y) * [1/2 * e^(x^2)] evaluated from 0 to 1
= e^(-y) * (1/2 * e^(1^2) - 1/2 * e^(0^2))
= e^(-y) * (1/2 * e - 1/2)
Therefore, the marginal density function of Y is:
f_Y(y) = e^(-y) * (1/2 * e - 1/2) for y > 0
= 0 otherwise
(c) Probability P(Y < 3X^2):
To calculate this probability, we need to integrate the joint density function over the region where Y < 3X^2.
P(Y < 3X^2) = ∫∫[y,1] 2xe^(x^2 - y) dx dy
We can rewrite the range of integration as:
y < 3x^2 for 0 < x < 1
0 < y < 1
Now we can evaluate the double integral:
P(Y < 3X^2) = ∫[0,1] ∫[0,y/3] 2xe^(x^2 - y) dx dy
Integrating 2xe^(x^2 - y) over x gives:
= e^(-y) * ∫[0,y/3] 2x e^(x^2) dx
= e^(-y) * [e^(x^2)] evaluated from 0 to y/3
= e^(-y) * (e^((y/3)^2) - e^0)
= e^(-y) * (e^(y^2/9) - 1)
Finally, we integrate this expression over y from 0 to 1:
P(Y < 3X^2) = ∫[0,1] e^(-y) * (e^(y^2/9) - 1) dy
This integral does not have a closed-form solution and needs to be evaluated numerically.
To find the marginal density function of X, we need to integrate the joint density function f(X, Y)(x, y) with respect to y over the entire range of y.
(a) Marginal Density Function of X:
To obtain the marginal density function of X, we integrate f(X, Y)(x, y) with respect to y from x^2 to infinity, since y > x^2.
∫[x^2, ∞] 2xe^(x^2 - y) dy
Integrating this expression gives us:
2xe^(x^2) ∫[x^2, ∞] e^(-y) dy
Since e^(-y) is an exponential function, we can integrate by using the property of exponential integrals:
∫ e^(-y) dy = -e^(-y)
Applying the limits of integration, the integral becomes:
2xe^(x^2) [-e^(-y)] evaluated from y = x^2 to y = ∞
Taking the difference between the two evaluations, we get:
2xe^(x^2) [-e^(-∞) - (-e^(-x^2))]
Simplifying, we have:
2xe^(x^2) [0 - (-e^(-x^2))]
2xe^(x^2) [e^(-x^2)]
The exponential term cancels out:
2x
Therefore, the marginal density function of X is:
f(X)(x) = 2x for 0 < x < 1
0 otherwise.
(b) Marginal Density Function of Y:
To find the marginal density function of Y, we integrate f(X, Y)(x, y) with respect to x over the entire range of x, which is from 0 to 1.
∫[0, 1] 2xe^(x^2 - y) dx
Integrating this expression gives us:
2 ∫[0, 1] xe^(x^2 - y) dx
To evaluate this integral, we can make a substitution u = x^2 - y, du = 2x dx:
∫ e^u du
Integrating, we get:
e^u
Substituting back, the integral becomes:
2 e^(x^2 - y)
Therefore, the marginal density function of Y is:
f(Y)(y) = 2e^(x^2 - y) for y > x^2
0 otherwise.
(c) Probability P(Y < 3X^2):
To compute this probability, we need to evaluate the integral of the joint density function f(X, Y)(x, y) with the condition Y < 3X^2.
∫[0,1] ∫[0,3x^2] 2xe^(x^2 - y) dy dx
Integrating with respect to y, we get:
∫[0,1] 2xe^(x^2 - y) evaluated from y = 0 to y = 3x^2
Simplifying, we have:
∫[0,1] 2xe^(x^2 - 3x^2) dx
∫[0,1] 2xe^(-2x^2) dx
This integral does not have a closed-form solution. However, you can numerically evaluate it using numerical integration techniques like the Trapezoidal Rule, Simpson's Rule, or using software like Mathematica or Python.
That completes the solution to the given problem.