1. ## Contiuous Uniform problem

Hi There,

I'm having troubles with this past paper question. I've touched on ont. Uniform stuff but this is really causing me trouble. Is the first question 0 and 1?

U is a continuous uniform random variable on the range (0, 1). A continuous random variable X is defined by X = 29(1 − U)−1/5.
1. Write down the largest and smallest values X can take.
2. Use the transformation of random variables technique to calculate the distribution function FX(x) of X.
3. Calculate the density function of X and the expectation and variance of X.
4. Suppose that X1, X2, ..., X10 are 10 independent copies of X. Use a suitable approximation to calculate the probability that the sample mean of these values is at least as large as 29.
5. Comment on the accuracy of the approximation.

2. 1. 0 and 1? Really? U is [0,1], then 1-U is on [1,0], then 29(1-U) is on [29,0], and finally 29(1-U)-(1/5) is on [29 - 1/5, 0 - 1/5]

Of course, you probably didn't mean what you wrote.

3. Originally Posted by dojo
Hi There,

I'm having troubles with this past paper question. I've touched on ont. Uniform stuff but this is really causing me trouble. Is the first question 0 and 1?

U is a continuous uniform random variable on the range (0, 1). A continuous random variable X is defined by X = 29(1 − U)−1/5.
1. Write down the largest and smallest values X can take.
2. Use the transformation of random variables technique to calculate the distribution function FX(x) of X.
3. Calculate the density function of X and the expectation and variance of X.
4. Suppose that X1, X2, ..., X10 are 10 independent copies of X. Use a suitable approximation to calculate the probability that the sample mean of these values is at least as large as 29.
5. Comment on the accuracy of the approximation.
What have you tried? Where are you stuck? Have you reviewed the 'transformation of random variables technique', how to calculate the mean and variance of the random variable aX + b, etc. ?

4. Originally Posted by TKHunny
1. 0 and 1? Really? U is [0,1], then 1-U is on [1,0], then 29(1-U) is on [29,0], and finally 29(1-U)-(1/5) is on [29 - 1/5, 0 - 1/5]

Of course, you probably didn't mean what you wrote.
Yes a typo indeed! should be (-1/5) as a power not a subtraction!!

i get x is U{29,0} whihc can't be right can it? The upper bound lower than th elower bound?

help.....

5. Originally Posted by dojo
yes a typo indeed! Should be (-1/5) as a power not a subtraction!!

I get x is u{29,0} whihc can't be right can it? The upper bound lower than th elower bound?

Help.....
$X=29(1-U)^{1/2}$

Well $V=1-U$ is also $\sim U(0,1)$ so we have:

$X=29(V)^{1/2}$

and as the square root function is increasing on $[0,1]$ the smallest value $X$ can take is $0$ and the targest is $29$.

CB

6. Thanks CaptainBlack - But the power is (-1/5) not 1/2 so I dont think this holds. I've thought some more and I think the smallest figure is 29 and the largest is +infinity.

What do you think?

7. Originally Posted by dojo
Thanks CaptainBlack - But the power is (-1/5) not 1/2 so I dont think this holds. I've thought some more and I think the smallest figure is 29 and the largest is +infinity.

What do you think?
Yes if the exponent is negative then you have a decreasing function and the range is 28 (the value at V=1) to infty (the value-iod at V=0)

CB