Function of continuous random variable?

2016-07-21 5:01 am

Find the pdf of Y if Y=X^2 and the pdf of X is given by f(x)=1, 0<x<1.

I found that f(y) = 1 * (1/2)(y^(-1/2)) = (1/2)(y^(-1/2))

But what is the interval for y, is it y ≥ 1?

回答 (1)

[匿名]

2016-07-21 7:23 am

Because 0 < X < 1, if Y = X² then 0 < Y < 1.
We call (0, 1) the support of Y, i.e., the range or interval you are asking for.

Your method directly applies the formula to obtain the pdf of the function of a random variable with an invertible transformation.

Here is another method.

Consider the cdf of Y
G(y)
= P(Y ≤ y)
= P(X² ≤ y)
= P(-√y ≤ X ≤ √y)
= P(0 ≤ X ≤ √y)
= ∫[0 ~ √y] 1 dx
= √y

Therefore, the pdf of Y is
g(y)
= d/dy G(y)
= d/dy √y
= 1/(2√y) for 0 < y < 1 and zero otherwise.

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