PDF of $sqrt X$ if $X$ has PDF $f_X(x) = x^3/4$ for $0 < x < 2$, otherwise $f_X(x) = 0$

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Let X have a density function $f_X(x) = x^3/4$ for 0 < x < 2, otherwise $f_X(x) = 0$

Let $Z = sqrtX$. Compute the density function of $f_Z(z)$ for $Z$.

---

My attempt:

Let $h(x) = sqrtx$, then $h^-1(z) = z^2$, and the derivative

$h'(x) = frac12x^1/2$

$$f_Z(z) = fracf_X(z^2)1/2z = fracz^6/41/2z = z^7/2$$ for $0 < z < sqrt2$, $0$ otherwise.

---

The solution in my textbook:

Since $Z^-1(z) = z$ and $Z'(z) = 1$, $f_Z(z) = f_X(z)(1) = z^3/4$.

Why is my answer wrong?

probability probability-distributions

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edited Jul 22 at 8:43

Did

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asked Jul 22 at 5:10

Bas bas

39611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the title of your textbook?
  – Did
  Jul 22 at 8:43
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

Let X have a density function $f_X(x) = x^3/4$ for 0 < x < 2, otherwise $f_X(x) = 0$

Let $Z = sqrtX$. Compute the density function of $f_Z(z)$ for $Z$.

---

My attempt:

Let $h(x) = sqrtx$, then $h^-1(z) = z^2$, and the derivative

$h'(x) = frac12x^1/2$

$$f_Z(z) = fracf_X(z^2)1/2z = fracz^6/41/2z = z^7/2$$ for $0 < z < sqrt2$, $0$ otherwise.

---

The solution in my textbook:

Since $Z^-1(z) = z$ and $Z'(z) = 1$, $f_Z(z) = f_X(z)(1) = z^3/4$.

Why is my answer wrong?

probability probability-distributions

share|cite|improve this question

edited Jul 22 at 8:43

Did

242k23208443

asked Jul 22 at 5:10

Bas bas

39611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the title of your textbook?
  – Did
  Jul 22 at 8:43
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Let X have a density function $f_X(x) = x^3/4$ for 0 < x < 2, otherwise $f_X(x) = 0$

Let $Z = sqrtX$. Compute the density function of $f_Z(z)$ for $Z$.

---

My attempt:

Let $h(x) = sqrtx$, then $h^-1(z) = z^2$, and the derivative

$h'(x) = frac12x^1/2$

$$f_Z(z) = fracf_X(z^2)1/2z = fracz^6/41/2z = z^7/2$$ for $0 < z < sqrt2$, $0$ otherwise.

---

The solution in my textbook:

Since $Z^-1(z) = z$ and $Z'(z) = 1$, $f_Z(z) = f_X(z)(1) = z^3/4$.

Why is my answer wrong?

probability probability-distributions

share|cite|improve this question

edited Jul 22 at 8:43

Did

242k23208443

asked Jul 22 at 5:10

Bas bas

39611

Let X have a density function $f_X(x) = x^3/4$ for 0 < x < 2, otherwise $f_X(x) = 0$

Let $Z = sqrtX$. Compute the density function of $f_Z(z)$ for $Z$.

---

My attempt:

Let $h(x) = sqrtx$, then $h^-1(z) = z^2$, and the derivative

$h'(x) = frac12x^1/2$

$$f_Z(z) = fracf_X(z^2)1/2z = fracz^6/41/2z = z^7/2$$ for $0 < z < sqrt2$, $0$ otherwise.

---

The solution in my textbook:

Since $Z^-1(z) = z$ and $Z'(z) = 1$, $f_Z(z) = f_X(z)(1) = z^3/4$.

Why is my answer wrong?

probability probability-distributions

share|cite|improve this question

edited Jul 22 at 8:43

Did

242k23208443

asked Jul 22 at 5:10

Bas bas

39611

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 8:43

Did

242k23208443

edited Jul 22 at 8:43

Did

242k23208443

edited Jul 22 at 8:43

Did

242k23208443

242k23208443

asked Jul 22 at 5:10

Bas bas

39611

asked Jul 22 at 5:10

Bas bas

39611

asked Jul 22 at 5:10

Bas bas

39611

39611

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the title of your textbook?
  – Did
  Jul 22 at 8:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the title of your textbook?
  – Did
  Jul 22 at 8:43
  

  
  
  
  

What is the title of your textbook?
– Did
Jul 22 at 8:43

What is the title of your textbook?
– Did
Jul 22 at 8:43

add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
down vote



accepted

I agree with your answer.
let $z in (0,sqrt2)$,
beginalign
Pr(Z le z) &= Pr(sqrtX le z)\
&= Pr(X le z^2)\
&=int_0^z^2f_X(t), dt
endalign

$$f_Z(z) = fracddz int_0^z^2f_X(t) , dt=f_X(z^2) (2z)=fracz^24(2z)=fracz^72$$

I don't see anywhere in the proposed solution that they use the property that $Z=sqrtX$ at all.

share|cite|improve this answer

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

add a comment | 

up vote
0
down vote

Your answer is correct. An alternative way to verify is to note that $$PrZ le z = PrsqrtX le z = PrX le z^2 = F_X(z^2).$$
Therefore,
$$f_Z(z) = fracddz F_X(z^2) = 2zf_X(z^2).$$

share|cite|improve this answer

answered Jul 22 at 5:17

Math Lover

12.4k21232

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

I agree with your answer.
let $z in (0,sqrt2)$,
beginalign
Pr(Z le z) &= Pr(sqrtX le z)\
&= Pr(X le z^2)\
&=int_0^z^2f_X(t), dt
endalign

$$f_Z(z) = fracddz int_0^z^2f_X(t) , dt=f_X(z^2) (2z)=fracz^24(2z)=fracz^72$$

I don't see anywhere in the proposed solution that they use the property that $Z=sqrtX$ at all.

share|cite|improve this answer

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

add a comment | 

up vote
0
down vote



accepted

I agree with your answer.
let $z in (0,sqrt2)$,
beginalign
Pr(Z le z) &= Pr(sqrtX le z)\
&= Pr(X le z^2)\
&=int_0^z^2f_X(t), dt
endalign

$$f_Z(z) = fracddz int_0^z^2f_X(t) , dt=f_X(z^2) (2z)=fracz^24(2z)=fracz^72$$

I don't see anywhere in the proposed solution that they use the property that $Z=sqrtX$ at all.

share|cite|improve this answer

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

I agree with your answer.
let $z in (0,sqrt2)$,
beginalign
Pr(Z le z) &= Pr(sqrtX le z)\
&= Pr(X le z^2)\
&=int_0^z^2f_X(t), dt
endalign

$$f_Z(z) = fracddz int_0^z^2f_X(t) , dt=f_X(z^2) (2z)=fracz^24(2z)=fracz^72$$

I don't see anywhere in the proposed solution that they use the property that $Z=sqrtX$ at all.

share|cite|improve this answer

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

I agree with your answer.
let $z in (0,sqrt2)$,
beginalign
Pr(Z le z) &= Pr(sqrtX le z)\
&= Pr(X le z^2)\
&=int_0^z^2f_X(t), dt
endalign

$$f_Z(z) = fracddz int_0^z^2f_X(t) , dt=f_X(z^2) (2z)=fracz^24(2z)=fracz^72$$

I don't see anywhere in the proposed solution that they use the property that $Z=sqrtX$ at all.

share|cite|improve this answer

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

answered Jul 22 at 5:17

Siong Thye Goh

77.6k134795

77.6k134795

add a comment | 

add a comment | 

up vote
0
down vote

Your answer is correct. An alternative way to verify is to note that $$PrZ le z = PrsqrtX le z = PrX le z^2 = F_X(z^2).$$
Therefore,
$$f_Z(z) = fracddz F_X(z^2) = 2zf_X(z^2).$$

share|cite|improve this answer

answered Jul 22 at 5:17

Math Lover

12.4k21232

add a comment | 

up vote
0
down vote

Your answer is correct. An alternative way to verify is to note that $$PrZ le z = PrsqrtX le z = PrX le z^2 = F_X(z^2).$$
Therefore,
$$f_Z(z) = fracddz F_X(z^2) = 2zf_X(z^2).$$

share|cite|improve this answer

answered Jul 22 at 5:17

Math Lover

12.4k21232

add a comment | 

up vote
0
down vote

up vote
0
down vote

Your answer is correct. An alternative way to verify is to note that $$PrZ le z = PrsqrtX le z = PrX le z^2 = F_X(z^2).$$
Therefore,
$$f_Z(z) = fracddz F_X(z^2) = 2zf_X(z^2).$$

share|cite|improve this answer

answered Jul 22 at 5:17

Math Lover

12.4k21232

Your answer is correct. An alternative way to verify is to note that $$PrZ le z = PrsqrtX le z = PrX le z^2 = F_X(z^2).$$
Therefore,
$$f_Z(z) = fracddz F_X(z^2) = 2zf_X(z^2).$$

share|cite|improve this answer

answered Jul 22 at 5:17

Math Lover

12.4k21232

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 5:17

Math Lover

12.4k21232

answered Jul 22 at 5:17

Math Lover

12.4k21232

answered Jul 22 at 5:17

Math Lover

12.4k21232

12.4k21232

add a comment | 

add a comment | 

 

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