# Some Questions in probability and statistics

• Jan 16th 2011, 10:31 AM
prosper
Some Questions in probability and statistics
1. Let X~Pois(1) what is the variance of X given x is bigger than 0?
a. (e-2)/(e-1)
b. e/(e-1)
c. 1/(e-2)
d. e(e-2)/(e-1)^2

I've tried to use the conditional variance formula but I had problems with calculating the
probability of x=k and x>0.

2. A worker calls his wife once a day. the length of each talk is independent of the others, its a random variable which distributes exponentially with a mean of 5.
One day the boss tells the worker that it is unacceptable and he has 3 more chances, if he calls his wife and talks for more than 10 minutes 3 times he will be fired.
What is the expected number of days the worker has before he gets fired?
a. 2e^3
b. 3e^3
c. 3e^(-2)
d. e^2 + e^3

The parameter of the exponential distribution is 1/5... this is easy.
but I have no idea what should I do next.
• Jan 16th 2011, 11:33 AM
Random Variable
$\displaystyle \displaystyle V(X|X>0) = E(X^2|X>0) - [E(X|X>0)]^2$

$\displaystyle \displaystyle E(X|X>0) = \frac{\sum_{x=1}^{\infty} xf(x)}{ P(X>0)}$

$\displaystyle \displaystyle E(X^{2}|X>0) = \frac{\sum_{x=1}^{\infty} x^{2}f(x)}{ P(X>0)}$

$\displaystyle \displaystyle f(x) = \frac{1}{x!} \frac{1}{e}$

$\displaystyle \displaystyle P(X>0) = 1 - P(X=0) = 1 - \frac{1}{e} = \frac{e-1}{e}$

$\displaystyle \displaystyle \sum_{x=1}^{\infty} xf(x) =\frac{1}{e} \sum^{\infty}_{x=1} \frac{x}{x!} = \frac{1}{e} \ e = 1$

$\displaystyle \displaystyle \sum_{x=1}^{\infty} x^{2}f(x) =\frac{1}{e} \sum^{\infty}_{x=1} \frac{x^{2}}{x!} = \frac{1}{e} \ 2e = 2$

so $\displaystyle \displaystyle E(X|X>0) = \frac{e}{e-1}$

and $\displaystyle \displaystyle E(X^{2}|X>0) = \frac{2e}{e-1}$

then finally $\displaystyle \displaystyle V(X|X>0) = \frac{2e}{e-1} - \frac{e^{2}}{(e-1)^{2}} = \frac{e(e-2)}{(e-1)^{2}}$
• Jan 16th 2011, 12:02 PM
prosper
Thanks! you've really helped me
any idea about the second question?
• Jan 17th 2011, 10:46 AM
prosper
some more questions I had trubles with
1. Let X1,X2 be independent random variables with equal distribution and CDF of:
$\displaystyle F_X (x) = { x \over x+1}, x \ge 0$
Let $\displaystyle Y=X1 \cdot X2$ the probability that Y is smaller or equal to 1 is:
a. 1/4
b. 1/2
c. 3/8
d. none of the above

I don't even know how to start solving this question....
• Jan 17th 2011, 03:02 PM
matheagle
You start by drawing.
you want to be "inside" the hyperbola xy<1, where x is x1 and y is x2.
By differentiation we have the marginals ...

$\displaystyle f(x)=(x+1)^{-2}$ where x>0

so the joint density is $\displaystyle f(x,y)= (x+1)^{-2} (y+1)^{-2}$ in the first quadrant

so you need to integrate $\displaystyle \int_0^{\infty}\int_0^{1/x}{dydx\over (x+1)^{2} (y+1)^{2}}$
• Jan 17th 2011, 03:48 PM
prosper
another question i ran into
Thanks.

1. The probability that a flip of a coin will get head is a continuous random variable P, with this PDF:
$\displaystyle f_P (p) = \left\{ \begin{array}{1 1} 2 & \quad {1 \over 4} \le p \le {3 \over 4} \\ 0 & \quad \mbox{otherwise} \\ \end{array} \right.}$
Let A be an event in which we got an head in the flip, what is the PDF of P given A?

another question which I solved but I'm not sure about the answer is:

2. What is the maximum estimator of
$\displaystyle P_X (x_i \theta) = c(\theta) e^{-\theta x} \quad ,x=0,1,... \quad ,\theta>0$
while $\displaystyle c(\theta)$ is a constant who dose not depend on x:
a. $\displaystyle \log \left( 1 + { n \over \sum_{i=1}^n x_i \right) }$
b. $\displaystyle \log \left( 1 + { \sum_{i=1}^n x_i \right) }$
c. $\displaystyle \log \left( n + { 1 \over \sum_{i=1}^n x_i \right) }$
d. none of the above.

I got d...
• Jan 17th 2011, 04:15 PM
matheagle
you left out something
I assume you want the MLE of theta, you estimate parameters.
It looks like you must solve for c, I get

$\displaystyle c(\theta)= 1-e^{-\theta}$ using the geometric series

so the likelihood function is

$\displaystyle L= \bigl(1-e^{-\theta}\bigr)^ne^{-\theta\sum x_i}$

Take the log, differentiate, then set equal to zero, flip it over and....THE answer is (a)
• Jan 17th 2011, 04:20 PM
prosper
yea.. i meant the MLE

Here is another one i solved:

Let $\displaystyle X~U(0,\theta)$ continuous. A confidence interval to $\displaystyle \theta$ is suggested based on a single observation:
the confidence level of the confidence interval is:
a. $\displaystyle 1-\alpha$
b. $\displaystyle 1-2\alpha$
c. $\displaystyle 1-{\alpha \over 2}$
d. you cannot calculate the confidence level from the given data

I've got b
• Jan 17th 2011, 04:34 PM
matheagle
I know, but MLE of which paramater, in this case it's obvious that it's theta
BUT you just don't ask for MLE's
It's the MLE of .....(an unknown parameter)
• Jan 17th 2011, 04:46 PM
prosper
I don't quite understand how you got (a) in the MLE question, especially how you found C could you please elaborate?

I've got another one. and I've added another one to the previous post i made.
I solved this one too the answer that i got is (a).

Let W~N(0,0.5). A sensor got signal W when hasnt penetration (H0) and signal W+Theta whan has penetration (alternative H1). Theta>0 constant.

Got record signal 0.96. Should we reject H0 with Statistical significance of 5% ?
a. Yes
b. No
c. The answer independ with theta