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Two more PMF/joint PMF questions

nacho

Active member
Sep 10, 2013
156
Please refer to the attached images.

For question 2,
How does this converge to one? i tried using a ratio test but thought it converged to zero

For question 3,
I understand how they get a $\frac{1}{1+b-a}$ in the denominator, but not how they set up the rest.
for example if Y = min(0,X)
then Y>0 is impossible because the min(0,X) restricts the value of Y to 0. Thus making $P(Y>0) = 0$
Similarly, $P(Y<a) = 0$ since the bounds are restricted from $a<0<b$.

for $y=0$ this can occur when X takes any value over $[0,b]$. There are $b+1$ such values.

However, for $a<y<0$, aren't there $0-a+1$ such values? Why do they simply have a $1$ in the numerator.
 

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chisigma

Well-known member
Feb 13, 2012
1,704
Please refer to the attached images.

For question 2,
How does this converge to one? i tried using a ratio test but thought it converged to zero
Remembering the well known expansion...

$\displaystyle e^{\lambda} = \sum_{n=0}^{\infty} \frac{\lambda^{n}}{n!}\ (1)$

... it is immediate to write...

$\displaystyle e^{- \lambda}\ \sum_{n=0}^{\infty} \frac{\lambda^{n}}{n!}= e^{-\lambda}\ e^{\lambda} = 1\ (2)$

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
For question 3,
I understand how they get a $\frac{1}{1+b-a}$ in the denominator, but not how they set up the rest.
for example if Y = min(0,X)
then Y>0 is impossible because the min(0,X) restricts the value of Y to 0. Thus making $P(Y>0) = 0$
Similarly, $P(Y<a) = 0$ since the bounds are restricted from $a<0<b$.

for $y=0$ this can occur when X takes any value over $[0,b]$. There are $b+1$ such values.

However, for $a<y<0$, aren't there $0-a+1$ such values? Why do they simply have a $1$ in the numerator.
If X is uniformely distributed in $a \le X \le b$ and is $a < 0$ then...


$\displaystyle P \{X = k \} = \frac{1}{1 + b - a}\ \forall k\ \text{with}\ a \le k \le b\ (1)$

If $Y = \text{min}\ (0,X)$ then is $\displaystyle P \{Y = k \} = \frac{1}{1 + b - a}\ \forall k\ \text{with}\ a \le k \le - 1$, $\displaystyle P \{Y = k \} = \frac{1+b}{1 + b - a}\ \text{if}\ k =0$ and 0 for all other k...


Kind regards


$\chi$ $\sigma$
 
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