Probability: Cumulative distribution problem

[tsamocki]

Homework Statement

Example 5.1.4: Y is a continuous random variable on the interval (0; 1) with
density function
fY (y) =
{3y2 0 < y < 1
{0 elsewhere
what is the cumulative distribution function of Y ?

Homework Equations

The relationship between a continuous random variable and the cumulative distribution function can be defined as: F(a) = P{X ∈ (-∞,α)} = ∫(-∞,α) f(x)dx

The Attempt at a Solution

Is the problem just asking for this: P{Y ∈ (0,1)} ?

In terms of the integration, the indefinite integral is y^3 + C.

Alas i am stuck!

 

[Ray Vickson]

Homework Statement

Example 5.1.4: Y is a continuous random variable on the interval (0; 1) with
density function
fY (y) =
{3y2 0 < y < 1
{0 elsewhere
what is the cumulative distribution function of Y ?

Homework Equations

The relationship between a continuous random variable and the cumulative distribution function can be defined as: F(a) = P{X ∈ (-∞,α)} = ∫(-∞,α) f(x)dx

The Attempt at a Solution

Is the problem just asking for this: P{Y ∈ (0,1)} ?

In terms of the integration, the indefinite integral is y^3 + C.

Alas i am stuck!

You wrote F(a) = ∫(-∞,α) f(x)dx. So, what is f(x) for x < 0? Can you replace the lower limit (-∞) by something else when a > 0? In other words, can you say
F(a) = \int_{\text{something}}^a f(x) \, dx, with 'something' ≠ -∞?

And NO: the problem is not asking you for P{Y ∈ (0,1)}; it is asking you for the cumulative distribution function (which you have already defined!).

RGV

 

[tsamocki]

You wrote F(a) = ∫(-∞,α) f(x)dx. So, what is f(x) for x < 0? Can you replace the lower limit (-∞) by something else when a > 0? In other words, can you say
F(a) = \int_{\text{something}}^a f(x) \, dx, with 'something' ≠ -∞?

And NO: the problem is not asking you for P{Y ∈ (0,1)}; it is asking you for the cumulative distribution function (which you have already defined!).

RGV

You mean this:

F(a) = \int_{\text{0}}^1 y^3 \, dx,

 

[tsamocki]

Edit:

F(a) = \int_{\text{0}}^1 3y^2 \, dy,

Sorry, i previously responded via phone.

 

[haruspex]

Edit:
F(a) = \int_{\text{0}}^1 3y^2 \, dy,

The RHS is not a function of a.