1. ## Exponential Distribution

The length of time Y necessary to complete a key operation in the construction has an exponential distribution with mean=10 hours. The formula C= 100 +40Y + 3Y^2 relates the cost C of completing this operation to the square of time to completion. Find the mean and variance of C.

Don't know where to start here =(.

2. Originally Posted by somestudent2
The length of time Y necessary to complete a key operation in the construction has an exponential distribution with mean=10 hours. The formula C= 100 +40Y + 3Y^2 relates the cost C of completing this operation to the square of time to completion. Find the mean and variance of C.

Don't know where to start here =(.
Start by making the moment-generating function of this distribution your friend.

Now note that C can be considered a linear function of 100, Y and Y^2. So:

1. E(C) = 100 + 40 E(Y) + 3 E(Y^2).

Get E(Y^2) from the moment generating function of Y - it's the second moment. Or just calculate it directly from the definition - in your case this reduces to finding $10 \int_{0}^{\infty} y^2 e^{-10y} \, dy$ (use repeated integration by parts).

2. Var(C) = 3^2 Var(Y^2) + 40^2 Var(Y) + 2(3)(40) Cov(Y^2, Y).

Note that Cov(Y^2, Y) = E(Y^3) - E(Y^2) E(Y).

Get E(Y^3) from the moment generating function of Y - it's the third moment. Or just calculate it directly from the definition - in your case this reduces to finding $10 \int_{0}^{\infty} y^3 e^{-10y} \, dy$ (use repeated integration by parts).

Details are left to you.

Alternatively you could just work out the pdf for C and then use it to find E(C) directly from the definition. I'd suggest the method of transformations since the pdf for Y is decreasing in y.

Then finding E(C^2) will enable you to calculate Var(C).

3. Thank you for the quick response. However i still feel pretty lost...

I don't quite understand how you produced those integrals.
When I solved the first integral i got 1, which is obviously not correct.

Also Var(C) = 3^2 Var(Y^2) + 40^2 Var(Y) + 2(3)(40) Cov(Y^2, Y)
what does cov stand for and why it is 3^2 40^2 and 2*3*40?

Since moment generating function for exponential distribution is (1-Bt)^-1, it confuses me
thanks again.

4. Originally Posted by somestudent2
Thank you for the quick response. However i still feel pretty lost...

I don't quite understand how you produced those integrals.

Also Var(C) = 3^2 Var(Y^2) + 40^2 Var(Y) + 2(3)(40) Cov(Y^2, Y)
what does cov stand for and why it is 3^2 40^2 and 2*3*40?

thanks again.
1. Are you familiar with the concept of covariance?

2. Are you familiar with the following definition?:

$E[g(Y)] = \int_{-\infty}^{\infty} g(y) \, f(y) \, dy$ , where g(Y) is a function of the random variable Y and f(y) is the pdf for the random variable Y.

So, for example, $E[Y^2] = \int_{-\infty}^{\infty} y^2 \, f(y) \, dy$.

5. yes I am familiar with the second concept but not with covariance.

6. Originally Posted by somestudent2
yes I am familiar with the second concept but not with covariance.
Well then, unless you get the pdf for the random variable C (using the technique I suggested, for example), I don't see how you can get the variance of C.

7. Hello again. I continued to work on this problem and looks like got through the first part. Got E(c)=104 3/50

I have a question on how to find Var(Y^2) and Var(Y)? From theorem i know that Var(Y)=E(Y^2) - u^2, what would be then Var(Y^2) and our u here?

would V(Y)=E(Y^2)-[E(Y)]^2

and V(Y^2)=E(Y^3)-[E(Y^2)]^2?

thanks

8. Originally Posted by somestudent2
Hello again. I continued to work on this problem and looks like got through the first part. Got E(c)=104 3/50

I have a question on how to find Var(Y^2) and Var(Y)? From theorem i know that Var(Y)=E(Y^2) - u^2, what would be then Var(Y^2) and our u here?

would V(Y)=E(Y^2)-[E(Y)]^2

and V(Y^2)=E(Y^3)-[E(Y^2)]^2? Mr F says: V(Y^2)=E([Y^2]^2)-[E(Y^2)]^2 = E(Y^4)-[E(Y^2)]^2.

thanks
..

9. ahhh i see, so I have to find the 4th moment as well, thanks