# Thread: Sum of two uniform distributions and other questions.

1. ## Sum of two uniform distributions and other questions.

Assume that
X has uniform distribution on the interval [0,1] and Y has uniform
distribution on the interval [0,2]. Find the density of Z = X + Y .

Would this just be 1+1/2 = 3/2?

Suppose X with values (0, 1) has density $\displaystyle $f(x) = cx^2(1-x)^2$$ for 0 < x < 1. Find:

a) the constant c; b) E[X]; c) Var[X]

I get, a) c = 30; b) 1/2; c) 1/28

Transistors produced by one machine have a lifetime which is exponentially distributed with mean 100 hours. Those produced by a second machine have an exponentially distributed lifetime with mean 200 hours. A package of 12 transistors contains 4 produced by the first machine, and 8 produced by the second machine. Let X be the lifetime of a transistor picked at random from this package. Find:

a) P(X greater or equal to 200 hours); b) E[X]; c) Var[X]

So would E[X] = (1/3)(100) + (2/3)(200) = 500/3? If so, then for a) I get e^(-6/5) = 0.3012, and for c) I get (500/3)^2 = 250000/9

2. Thats too many questions in a single thread. Try to post only one question in a thread. Its easy to read and makes your thread less messy.

1) I dont think so. Look at this link(page No 8) that finds the pdf of sum of 2 independent uniform RVs.

2)the values look correct.

Edit: No 3.. Your expected value is correct.. not the variance.. You are doing $\displaystyle V(X)=(E(X))^2$, which is not correct.

You have found $\displaystyle E[X]$..then you need to find $\displaystyle E[X^2]$ (using conditional expectation) and the variance is found the following identity:

$\displaystyle V(X)=E(X^2)-E^2(X)$

3. But for exponential r.v.'s isn't E[X]=SD[X]? So then to get variance, couldn't you just square SD[X]?

4. the sum of two indep uniforms is triangular
the easiest technique to obtain the CDF of the sum via geometry on the unit square

5. I got the sum of the RVs, thanks!

Could someone verify the third question? Is the logic correct in my previous post?

6. what is the third question?
The density $\displaystyle x^2(1-x)^2$ on 0<x<1 is a Beta, so all those questions are obvious.

7. Originally Posted by BrownianMan
Assume that
[LEFT]X has uniform distribution on the interval [0,1] and Y has uniform
distribution on the interval [0,2]. Find the density of Z = X + Y .

Would this just be 1+1/2 = 3/2?
In general if $\displaystyle \phi_{x} (*)$ is the PDF of a random variable x and $\displaystyle \phi_{y} (*)$ is the PDF of a random variable y , the PDF od the random variable z=x+y is given by...

$\displaystyle \displaystyle \phi_{z} (z) = \phi_{x} (z) * \phi_{y} (z)$ (1)

... where * means 'convolution'. The (1) permits to use fuitfully L-transform od F- transform tecniques to obtain in confortable way the $\displaystyle \phi_{z} (*)$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. Originally Posted by chisigma
In general if $\displaystyle \phi_{x} (*)$ is the PDF of a random variable x and $\displaystyle \phi_{y} (*)$ is the PDF of a random variable y , the PDF od the random variable z=x+y is given by...

$\displaystyle \displaystyle \phi_{z} (z) = \phi_{x} (z) * \phi_{y} (z)$ (1)

... where * means 'convolution'. The (1) permits to use fuitfully L-transform od F- transform tecniques to obtain in confortable way the $\displaystyle \phi_{z} (*)$...
In the problem proposed by BrownianMan is...

$\displaystyle \displaystyle \mathcal{L} \{\phi_{x} (*)\} = \Phi_{x} (s) = \frac{1-e^{-s}}{s}$

$\displaystyle \displaystyle \mathcal{L} \{\phi_{y} (*)\} = \Phi_{y} (s) = \frac{1-e^{-2 s}}{2 s}$ (1)

... so that is...

$\displaystyle \displaystyle \phi_{z} (z)= \mathcal{L}^{-1}\{\frac{1-e^{-s} - e^{-2 s} + e^{-3 s}}{2 s^{2}} \}$ (2)

The function $\displaystyle \phi_{z} (z)$ is illustrated here...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

9. this is the third question:

Transistors produced by one machine have a lifetime which is exponentially distributed with mean 100 hours. Those produced by a second machine have an exponentially distributed lifetime with mean 200 hours. A package of 12 transistors contains 4 produced by the first machine, and 8 produced by the second machine. Let X be the lifetime of a transistor picked at random from this package. Find:

a) P(X greater or equal to 200 hours); b) E[X]; c) Var[X]

So would E[X] = (1/3)(100) + (2/3)(200) = 500/3? If so, then for a) I get e^(-6/5) = 0.3012, and for c) I get (500/3)^2 = 250000/9

It is independent of the others.

10. As told in post No 2, the expected value is correct, but the variance is wrong.