Math Help - Do I need to use U substitution?

1. Do I need to use U substitution?

I'm trying to integrate
$f(y) = k (1-y)^4$.

Y is a random variable, K is a constant.

I think I need to do this because it is a density function and I need a distribution function.

I decide to give it a shot without U substituting, because I don't know if need it and I certainly haven't got the hang of it yet. (I know I have to U substitute if there is anything in the equation other than X, but I gather I can take out coefficients like k, so I'm not sure about this one.)

I decided to let 1-y = A. So:
$\int k (1-y)^4 = k A^5/5 + C$

Moving from the antiderivative to the definite integral is where it goes bad. I don't think I've ever gotten a negative answer to a definite integral, but on the interval (0, 0.4) I get -0.184k. [Edit: This is what I got in the solution below; in both cases my conditional probability is 2.25.]

Assuming I do need to use U substitution, does anyone have any suggestions on pointers for how to use it? The book I'm using is getting worn with my re-reading the pair of brief sentences which are supposed to explain it....

2. Originally Posted by Boris B
I'm trying to integrate
$f(y) = k (1-y)^4$.

Y is a random variable, K is a constant.

I think I need to do this because it is a density function and I need a distribution function.

I decide to give it a shot without U substituting, because I don't know if need it and I certainly haven't got the hang of it yet. (I know I have to U substitute if there is anything in the equation other than X, but I gather I can take out coefficients like k, so I'm not sure about this one.)

I decided to let 1-y = A. So:
$\int k (1-y)^4 = k A5^5 + C$

Moving from the antiderivative to the definite integral is where it goes bad. I don't think I've ever gotten a negative answer to a definite integral, but on the interval (0, 0.4) I get -0.184k.

Assuming I do need to use U substitution, does anyone have any suggestions on pointers for how to use it? The book I'm using is getting worn with my re-reading the pair of brief sentences which are supposed to explain it....
if
$u=(1-y) \mbox{ then } du=-dy$

You forgot the minus sign

3. Okay, I've set
u = 1-y
du = -1dy

-du = dy

So
$\int k (1-y)^4 dy = k \int u^4 \cdot -du$
$-k 4 u^3 du$

The definite integrals are still zero though. Even if I assume there shouldn't be a negative in front of the final k, my larger integral is still smaller than the one that should be larger. (The rest of the problem has me comparing $\int_0^{0.4} k (1-y)^4 dy$ to $\int_0^{0.1} k (1-y)^4 dy$. The latter should be quite a bit larger but it is much smaller.)

I guess I'm still not really comfortable with u substitution. It seems like dy (or dx or whatever, from the original equation) was just a kind of marker meaning "with respect to" that trailed the problem. After substituting it for du, du become a factor in the original sense, something we have to multiply the rest of the problem by. Since it's okay in math to string factors together without multiplicaton signs, intending them to be multiplied, it's hard to say when du or dx or whatever is to be multiplied and when it just means "with respect to". I kind of miss multiplication signs.

4. Originally Posted by Boris B
Okay, I've set
u = 1-y
du = -1dy

-du = dy

So
$\int k (1-y)^4 dy = k \int u^4 \cdot -du$
$-k 4 u^3 du$

The definite integrals are still zero though. Even if I assume there shouldn't be a negative in front of the final k, my larger integral is still smaller than the one that should be larger. (The rest of the problem has me comparing $\int_0^{0.4} k (1-y)^4 dy$ to $\int_0^{0.1} k (1-y)^4 dy$. The latter should be quite a bit larger but it is much smaller.)

I guess I'm still not really comfortable with u substitution. It seems like dy (or dx or whatever, from the original equation) was just a kind of marker meaning "with respect to" that trailed the problem. After substituting it for du, du become a factor in the original sense, something we have to multiply the rest of the problem by. Since it's okay in math to string factors together without multiplicaton signs, intending them to be multiplied, it's hard to say when du or dx or whatever is to be multiplied and when it just means "with respect to". I kind of miss multiplication signs.
I think you took the derivative instead of the integral.

Also see the graphs below for the area.

$\int k(1-y)^4dy$

let $u=1-y \mbox{ then } du=-dy \iff -du=dy$

$k\int u^4 (-du)=-k \int u^4 du=-\frac{k}{5}u^5=-\frac{k}{5}(1-y)^5$

from 0 to 0.1

from 0 to 0.4

I hope this helps

5. "I think you took the derivative instead of the integral."

I can't believe I climbed the whole mountain only two trip over my oxygen cannister (metaphor strained due to altitude sickness).

6. Originally Posted by Boris B
I'm trying to integrate
$f(y) = k (1-y)^4$.

Y is a random variable, K is a constant.

I think I need to do this because it is a density function and I need a distribution function.

I decide to give it a shot without U substituting, because I don't know if need it and I certainly haven't got the hang of it yet. (I know I have to U substitute if there is anything in the equation other than X, but I gather I can take out coefficients like k, so I'm not sure about this one.)

I decided to let 1-y = A. So:
$\int k (1-y)^4 = k A5^5 + C$

Moving from the antiderivative to the definite integral is where it goes bad. I don't think I've ever gotten a negative answer to a definite integral, but on the interval (0, 0.4) I get -0.184k.

Assuming I do need to use U substitution, does anyone have any suggestions on pointers for how to use it? The book I'm using is getting worn with my re-reading the pair of brief sentences which are supposed to explain it....
To get the distribution function you need to include the integral terminals .... over what domain is f(y) = k (1-y)^4 ?

Or are you integrating to find the value of k?

It would be beneficial if you gave the entire question, not just a small isolated portion that has no context.

7. Whole Question

"A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by V = 100,000Y
where Y is a random variable with density function
$f(y)= k(1-y)^4$ for 0<y<1
$f(y)= 0$ otherwise
where k is a constant.
What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000?"

From there I just sort of quit using V and compared the intervals (0, 0.1) and (0, 0.4). I have no idea if that is right. As far as I know, I'll have to divide the probability of V exceeding 40,000 by the probability that it exceeds 10,000, which is why I want the latter to be larger.

8. Originally Posted by Boris B
Having taken the antiderivative of $\int k (1-y)^4 dy$, I came up with $k x^5 / 5$ using u substitution (u = 1-y, du = -dy). This is probably wrong, because
Two things. You set up your u-sub correctly but I think you're missing a negative sign. It should be $-\frac{ku^5}{5}$. And did you remember to plug back in 1-y for u?

9. I did remember to plug (1-y) back in, but something has gone astray, because I am back with the answer I had before I knew how to do U substitution:
P (y>0.4 | y>0.1) = 2.25.

$\int k (1-y)^4 = k \int u^4 -du$
$k \int u^4 -du = -k/5 u^5$
$-k/5 u^5 = -k/5 (1-y)^5$

When
y=0.4, $-k/5 (1-y)^5 = -0.0778 k/5$
y=0.1, $-k/5 (1-y)^5 = -0.5905 k/5$
y=0, $-k/5 (1-y)^5 = -1 k/5$
So
$\int k_0^{0.4} (1-y)^4 = -0.9222k / 5$
$\int k_0^{0.1} (1-y)^4 = -0.4095k / 5$
Dividing them means the -ks cancel, so I'm left with a probability exceeding 1.

I wonder if I have the intervals backwards. I wonder if I should be using
$\int k_{0.4}^0 (1-y)^4$, etc.

10. I don't really understand the problem you're trying to solve. I see you're doing some definite integration, but could you post the problem you're having trouble with?

EDIT: Ok I read your OP. I'm still not clear on everything, but perhaps check your bounds?

11. Originally Posted by Boris B
"A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by V = 100,000Y
where Y is a random variable with density function
$f(y)= k(1-y)^4$ for 0<y<1
$f(y)= 0$ otherwise
where k is a constant.
What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000?"

From there I just sort of quit using V and compared the intervals (0, 0.1) and (0, 0.4). I have no idea if that is right. As far as I know, I'll have to divide the probability of V exceeding 40,000 by the probability that it exceeds 10,000, which is why I want the latter to be larger.
You require

$\Pr(V > 40,000 | V > 10,000) = \Pr(Y > 0.4 | Y > 0.1)$

$= \frac{\Pr(Y > 0.4 \, \text{and} \, Y > 0.1)}{\Pr(Y > 0.1)} = \frac{\Pr(Y > 0.4)}{\Pr(Y > 0.1)}$.

$\Pr(Y > 0.4) = \int_{0.4}^{1} k(1-y)^4 \, dy = \frac{243 k}{15,625}$.

$\Pr(Y > 0.1) = \int_{0.4}^{1} k(1-y)^4 \, dy = \frac{59,049 k}{500,000}$.

Therefore $\Pr(V > 40,000 | V > 10,000) = \frac{243/15,625}{59,049/500,000} = \frac{32}{243} \approx 0.131687$.

If you're studying pdf's it is assumed that you can integrate. You must go back and thoroughly revise basic integration techniques. I'll do $\int_{0.4}^{1} k(1-y)^4 \, dy$ in detail:

Make the substitution u = 1 - y. Then:

$\frac{du}{dy} = -1 \Rightarrow dy = - du$.

$y = 0.4 = \frac{2}{5} \Rightarrow u = 1 - \frac{2}{5} = \frac{3}{5}$.

$y = 1 \Rightarrow u = 0$.

So the integral becomes

$\int_{3/5}^{0} k u^4 \, (-du) = - k \int_{3/5}^{0} u^4 \, du = k \int_{0}^{3/5} u^4 \, du = \frac{k}{5} u^5 \bigg{|}_{0}^{3/5} = \frac{k}{5} \, \left( \frac{3}{5} \right)^5 = \frac{243k}{15,625}$.

12. I think I understood everything but this.

$\Pr(Y > 0.1) = \int_{0.4}^{1} k(1-y)^4 \, dy = \frac{59,049 k}{500,000}$

Are the numbers on the integral right? I'm hoping it's a typo, or I'm really confused.

13. Ok

Originally Posted by Boris B
I think I understood everything but this.

$\Pr(Y > 0.1) = \int_{0.4}^{1} k(1-y)^4 \, dy = \frac{59,049 k}{500,000}$

Are the numbers on the integration right? I'm hoping it's a typo, or I'm really confused.
I have not read all the prior info but the answer should be $\frac{-k(1-y)^5}{5}\bigg|_{.4}^{1}$

14. Originally Posted by Boris B
I think I understood everything but this.

$\Pr(Y > 0.1) = \int_{0.4}^{1} k(1-y)^4 \, dy = \frac{59,049 k}{500,000}$

Are the numbers on the integral right? I'm hoping it's a typo, or I'm really confused.
Clearly it's a typo.

The correction is:

$
\Pr(Y > 0.1) = \int_{{\color{red}0.1}}^{1} k(1-y)^4 \, dy = \frac{59,049 k}{500,000}
$

15. I got the same answer you did. I had put 0 in the lower bounds instead of 0.1 and 0.4, and 0.1 and 0.4 in the upper bounds instead of 1. So even the calculations that I did correctly led me nowhere.

I feel much better now.