Probability density function of uniform distribution

[diracdelta]

Homework Statement

Random variable X is uniformly distributed on interval [0,1]:
$$f(x)=\begin{cases} 1 & \text{ if } 0\leq x\leq 1\\ 0 & \text{ else} \end{cases}$$

a) Find probability density function ρ(y) of random variable $$Y=\sqrt{X} +1$$

I tried like this. Is it good, if no why not?

$$F_{Y}(y)=P(Y\leq y)=P(x^{2}+1\leq y)=P(x\leq y^{2}+1)=F_{x}(y^{2}+1) \\ f_{X}(x)=F_{X}'(y^{2}+1)=f_{X}(y^{2}+1)* \left | \frac{d}{dy} (y^{2}+1)\right |=f_{X}(y^{2}+1)*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$

 

[HallsofIvy]

Homework Statement

Random variable X is uniformly distributed on interval [0,1]:
$$f(x)=\begin{cases} 1 & \text{ if } 0\leq x\leq 1\\ 0 & \text{ else} \end{cases}$$

a) Find probability density function ρ(y) of random variable $$Y=\sqrt{X} +1$$

I tried like this. Is it good, if no why not?

$$F_{Y}(y)=P(Y\leq y)=P(x^{2}+1\leq y)=P(x\leq y^{2}+1)=F_{x}(y^{2}+1) \\ f_{X}(x)=F_{X}'(y^{2}+1)=f_{X}(y^{2}+1)* \left | \frac{d}{dy} (y^{2}+1)\right |=f_{X}(y^{2}+1)*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$

How did $\sqrt{X}+ 1$ become $X^2+ 1$? If you intended the inverse function then it should be $X= (Y- 1)^2$

 

[diracdelta]

Hmm, i see.
Ok then, i get this
$$F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2} \\ f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |=f_{X}(y+1)^{2})*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$[/QUOTE]

But what about y's probability density function ?
Is it ok?

 

[haruspex]

$F_{Y}(y)=P(Y\leq y)=P(x^{\frac 12}+1\leq y)$

That should really be
$F_{Y}(y)=P(Y\leq y)=P(X^{\frac 12}+1\leq y)$

$=P(x\leq (y+1)^{2})$

Check that step.

$f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |$

$F_{X}'((y+1)^{2})$ just means the derivative wrt x of F_X(x) evaluated at x = (y+1)²
$f_{X}(x)=F_{X}'((y+1)^{2})=f_{X}((y+1)^{2})$

 

[Ray Vickson]

Hmm, i see.
Ok then, i get this
$$F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2} \\ f_{X}(x)=F_{X}'(y+1)^{2}=f_{X}(y+1)^{2}* \left | \frac{d}{dy} (y+1)^{2})\right |=f_{X}(y+1)^{2})*2y \\ f_{Y}(y)=\begin{cases} 2y & \text{ if } 1\leq x\leq 2\\ 0 & \text{ else} \end{cases}$$

But what about y's probability density function ?
Is it ok?[/QUOTE]

How do you get from $x^{1/2}+1 \leq y$ to $x \leq (y+1)^2$? (I don't.)

Anyway, since $X$ ranges from 0 to 1, so does $\sqrt{X}$ and that means that $Y = 1 + \sqrt{X}$ ranges from 1 to 2. So, you ought to have $F_Y(1) = 0$ and $F_Y(2) = 1$. Do you have that?

 

[andrewkirk]

$$F_{Y}(y)=P(Y\leq y)=P(x^{1/2}+1\leq y)=P(x\leq (y+1)^{2}) =F_{x}(y+1)^{2}$$

Up to here it's OK, except for the two errors pointed out by haruspex. Fix those and you will have a correct formula for $F_Y(y)$ in terms of $F_X$ applied to some function of $y$.

Next you can use the fact that $F_X(x)=\int_0^x f_X(u)du=\int_0^x du=x$ to write a formula for $F_{Y}(y)$ in terms of $y$.

Forget about everything you wrote below your first line, as that heads off on a goose chase involving $f_X$ and there is no need to refer to $f_X$ any further.

Differentiating your formula for $F_Y(y)$ will give you a formula for $f_Y(y)$.

To complete the answer you need to put range constraints into your expression for $f_Y$, along the lines suggested by Ray.