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# How to start solving?

Let Y ~ U(0,1).

a. Derive the distribution of the random variable W = Y2.

b. Derive the distribution of the random variable W = square root of Y.

### 2 Answers

- Astral WalkerLv 71 decade agoFavorite Answer
Assuming you mean W=Y²

Let FY(y) be the cumulative distribution function of random variable Y and FW(w) be the cdf of random variable W=f(Y). We can then say that

FY(y) = P[X<x]

FW(w) = P[f(X) < w]

Now let W=Y²

FW(w) = P[Y² < w]

P(Y²<w) = P(|Y|<√w) = P(-√w < Y < √w)

so

FW(w) = FY(√w) - FY(-√w)

Since Y ~ U(0,1) then FY(x) = x

FW(w) = FY(√w) - FY(-√w)

FW(w) = √w + √w

FW(w) = 2√w

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Now let W=√Y

FW(w) = P[√Y < w]

P(√Y<w) = P(Y<w²)

so

FW(w) = FY(w²)

Since Y ~ U(0,1) then FY(x) = x

FW(w) = FY(w²)

FW(w) = w²

- Scarlet ManukaLv 71 decade ago
Y has pdf f(x) = 1, 0 ≤ x ≤1.

Let the pdf of W = F(Y) be g(x). We need to have

g(x) Δx = ∫ f(x) dx where the integral is over the set A = {x: g(x) <= F(x) <= g(x) + Δx}

If F is invertible on the domain of Y, this will be over the interval (F^-1(x), F^-1(x) + δx) or (F^-1(x) + δx, F^-1(x)) (depending on whether δx is positive or negative) where δx F'(x) = Δx. So we get

g(x) Δx = f(F^-1(x)) Δx / |F'(F^-1(x))|

and hence we get the rule

g(x) = f(F^-1(x)) / |F'(F^-1(x))|

or, in a more easily understandable form,

g(F(x)) = f(x) / |F'(x)|

But remember this only applies when F is invertible; otherwise we have to add the contributions from each part of the integral corresponding to the same value of F. In this question both functions are invertible on [0, 1].

So we get:

a) domain of g(x) is also [0, 1], g(x) = 1 / 2x

b) domain of g(x) is also [0, 1], g(x) = 1 / ((1/2) x^(-1/2)) = 2√x.