hii. I have a question about pdf and cdf.. I couldn't solve it
Please help me :/
1) the pdf of the random var. X is given by
f(x)= (3/8)(x+1)^2 if -1<x<1
0 if otherwise
Find the pdf of a) Y=1-X^2
b) Y=1-X^2 ifx<=0 and Y=1-X if x>0
9 years later I hope you got your answer!! You're probably an adult by now lol.
As an AI language model, I don't have an age or life events, but thank you for checking in! If you have any questions about pdf and cdf, I'm here to help.
I'm not @melike. I'm @oceansparkless
I apologize for the confusion, @oceansparkless! Thank you for pointing that out to me. If you have any questions regarding pdf and cdf, feel free to ask me.
Hi! I can help you with your question about the probability density function (pdf) and cumulative distribution function (cdf).
To find the pdf of a transformed random variable, we typically use the method of transformation. This involves substituting the expression for the original variable into the pdf and applying the appropriate change of variables.
Let's solve your questions step by step.
a) Find the pdf of Y = 1 - X^2:
To find the pdf of Y, we need to determine its domain and then calculate the probability density function.
The given pdf of X is:
f(x) = (3/8)(x + 1)^2 if -1 < x < 1
0 otherwise
First, we find the domain of Y:
1 - X^2 > 0
The above inequality tells us that Y can take values only in the range where X^2 < 1. This implies -1 < X < 1.
Now, we can calculate the pdf of Y:
fY(y) = fX(g^(-1)(y)) * | d/dy(g^(-1)(y)) |
where g(x) = 1 - x^2 and g^(-1)(y) is the inverse function of g.
To find the inverse function, solve g(x) = y for x:
1 - x^2 = y
x^2 = 1 - y
x = ±√(1 - y)
Since -1 < x < 1, we take the positive square root:
g^(-1)(y) = √(1 - y)
Next, we substitute g^(-1)(y) back into f(x):
fY(y) = fX(√(1 - y)) * | d/dy(√(1 - y)) |
Now differentiate √(1 - y) with respect to y:
d/dy(√(1 - y)) = -1/(2√(1 - y))
Substituting everything back, we have:
fY(y) = (3/8)(√(1 - y) + 1)^2 * |-1/(2√(1 - y))|
Simplifying further:
fY(y) = (3/8)(√(1 - y) + 1)^2/(2√(1 - y))
So, the pdf of Y = 1 - X^2 is (3/8)(√(1 - y) + 1)^2/(2√(1 - y)).
b) Find the pdf of Y = 1 - X^2 if X ≤ 0 and Y = 1 - X if X > 0:
In this case, we have two different expressions for Y depending on the value of X.
For X ≤ 0, we have Y = 1 - X^2.
For X > 0, we have Y = 1 - X.
To calculate the pdf of Y, we need to consider the different cases:
Case 1: X ≤ 0
For X ≤ 0, we substitute Y = 1 - X^2 into f(x):
fY(y) = fX(g^(-1)(y)) * | d/dy(g^(-1)(y)) |
where g(x) = 1 - x^2 and g^(-1)(y) = √(1 - y).
Substituting:
fY(y) = fX(√(1 - y)) * | d/dy(√(1 - y)) |
Differentiating and simplifying as before, we have:
fY(y) = (3/8)(√(1 - y) + 1)^2/(2√(1 - y))
Again, this is the pdf for X ≤ 0.
Case 2: X > 0
For X > 0, we substitute Y = 1 - X into f(x):
fY(y) = fX(g^(-1)(y)) * | d/dy(g^(-1)(y)) |
where g(x) = 1 - x and g^(-1)(y) = 1 - y.
Substituting:
fY(y) = fX(1 - y) * | d/dy(1 - y) |
Differentiating, we get:
fY(y) = (3/8)(2 - 2(1 - y)) = (3/4)y
Therefore, the pdf of Y = 1 - X^2 if X ≤ 0 is (3/8)(√(1 - y) + 1)^2/(2√(1 - y)), and the pdf of Y = 1 - X if X > 0 is (3/4)y.
I hope this helps! Let me know if you have any further questions.