Probability function- Random variable [on hold]
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I need to find a Probability function using
$$
F(x) = begin{cases}
0 & text{if $x<0$}\
frac{1-a^k}{1-a^{n+1}} & text{if $k-1 leq x leq k, k=1,dots,n, n>1$}\
1 & text{if $x geq n$}
end{cases}
$$
Could someone help me with this exercise?
Thank you in advance.
probability probability-distributions random-variables
edited Nov 18 at 11:58
amWhy
191k27223439
asked Nov 18 at 11:47
Francisco
82
put on hold as unclear what you're asking by NCh, ancientmathematician, DRF, Cesareo, John B yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
I need to find a Probability function using
$$
F(x) = begin{cases}
0 & text{if $x<0$}\
frac{1-a^k}{1-a^{n+1}} & text{if $k-1 leq x leq k, k=1,dots,n, n>1$}\
1 & text{if $x geq n$}
end{cases}
$$
Could someone help me with this exercise?
Thank you in advance.
probability probability-distributions random-variables
edited Nov 18 at 11:58
amWhy
191k27223439
asked Nov 18 at 11:47
Francisco
82
put on hold as unclear what you're asking by NCh, ancientmathematician, DRF, Cesareo, John B yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$F(x)$ has the properties of a distribution function as long as $0lt alt 1$. Define what you are looking for. (Probability function?)
– herb steinberg
Nov 18 at 20:13
Yes, Sorry I forgot to write that: 0 < a < 1. I need to find the Probability function of the corresponding cdf.
– Francisco
Nov 18 at 21:56
My book calls it Probability function. The result is: p(x)= (a^x)/[sum k=1,..,n of (a^k)] for x=0,..,n
– Francisco
Nov 19 at 6:03
$frac{a^k-a^{k+1}}{1-a^{n+1}}=a^kfrac{1-a}{1-a^{n+1}}=frac{a^k}{sum_{j=0}^na^j}$, so we agree. (Almost, check lower limit of sum).
– herb steinberg
Nov 19 at 16:50
Since there is a jump at x=0, you need to include k=0 in the sum.
– herb steinberg
Nov 19 at 22:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to find a Probability function using
$$
F(x) = begin{cases}
0 & text{if $x<0$}\
frac{1-a^k}{1-a^{n+1}} & text{if $k-1 leq x leq k, k=1,dots,n, n>1$}\
1 & text{if $x geq n$}
end{cases}
$$
Could someone help me with this exercise?
Thank you in advance.
probability probability-distributions random-variables
edited Nov 18 at 11:58
amWhy
191k27223439
asked Nov 18 at 11:47
Francisco
82
I need to find a Probability function using
$$
F(x) = begin{cases}
0 & text{if $x<0$}\
frac{1-a^k}{1-a^{n+1}} & text{if $k-1 leq x leq k, k=1,dots,n, n>1$}\
1 & text{if $x geq n$}
end{cases}
$$
Could someone help me with this exercise?
Thank you in advance.
probability probability-distributions random-variables
probability probability-distributions random-variables
edited Nov 18 at 11:58
amWhy
191k27223439
asked Nov 18 at 11:47
Francisco
82
edited Nov 18 at 11:58
amWhy
191k27223439
asked Nov 18 at 11:47
Francisco
82
edited Nov 18 at 11:58
amWhy
191k27223439
edited Nov 18 at 11:58
amWhy
191k27223439
edited Nov 18 at 11:58
amWhy
191k27223439
191k27223439
asked Nov 18 at 11:47
Francisco
82
asked Nov 18 at 11:47
Francisco
82
asked Nov 18 at 11:47
Francisco
82
82
put on hold as unclear what you're asking by NCh, ancientmathematician, DRF, Cesareo, John B yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by NCh, ancientmathematician, DRF, Cesareo, John B yesterday
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$F(x)$ has the properties of a distribution function as long as $0lt alt 1$. Define what you are looking for. (Probability function?)
– herb steinberg
Nov 18 at 20:13
Yes, Sorry I forgot to write that: 0 < a < 1. I need to find the Probability function of the corresponding cdf.
– Francisco
Nov 18 at 21:56
My book calls it Probability function. The result is: p(x)= (a^x)/[sum k=1,..,n of (a^k)] for x=0,..,n
– Francisco
Nov 19 at 6:03
$frac{a^k-a^{k+1}}{1-a^{n+1}}=a^kfrac{1-a}{1-a^{n+1}}=frac{a^k}{sum_{j=0}^na^j}$, so we agree. (Almost, check lower limit of sum).
– herb steinberg
Nov 19 at 16:50
Since there is a jump at x=0, you need to include k=0 in the sum.
– herb steinberg
Nov 19 at 22:52
add a comment |
$F(x)$ has the properties of a distribution function as long as $0lt alt 1$. Define what you are looking for. (Probability function?)
– herb steinberg
Nov 18 at 20:13
Yes, Sorry I forgot to write that: 0 < a < 1. I need to find the Probability function of the corresponding cdf.
– Francisco
Nov 18 at 21:56
My book calls it Probability function. The result is: p(x)= (a^x)/[sum k=1,..,n of (a^k)] for x=0,..,n
– Francisco
Nov 19 at 6:03
$frac{a^k-a^{k+1}}{1-a^{n+1}}=a^kfrac{1-a}{1-a^{n+1}}=frac{a^k}{sum_{j=0}^na^j}$, so we agree. (Almost, check lower limit of sum).
– herb steinberg
Nov 19 at 16:50
Since there is a jump at x=0, you need to include k=0 in the sum.
– herb steinberg
Nov 19 at 22:52
$F(x)$ has the properties of a distribution function as long as $0lt alt 1$. Define what you are looking for. (Probability function?)
– herb steinberg
Nov 18 at 20:13
$F(x)$ has the properties of a distribution function as long as $0lt alt 1$. Define what you are looking for. (Probability function?)
– herb steinberg
Nov 18 at 20:13
Yes, Sorry I forgot to write that: 0 < a < 1. I need to find the Probability function of the corresponding cdf.
– Francisco
Nov 18 at 21:56
Yes, Sorry I forgot to write that: 0 < a < 1. I need to find the Probability function of the corresponding cdf.
– Francisco
Nov 18 at 21:56
My book calls it Probability function. The result is: p(x)= (a^x)/[sum k=1,..,n of (a^k)] for x=0,..,n
– Francisco
Nov 19 at 6:03
My book calls it Probability function. The result is: p(x)= (a^x)/[sum k=1,..,n of (a^k)] for x=0,..,n
– Francisco
Nov 19 at 6:03
$frac{a^k-a^{k+1}}{1-a^{n+1}}=a^kfrac{1-a}{1-a^{n+1}}=frac{a^k}{sum_{j=0}^na^j}$, so we agree. (Almost, check lower limit of sum).
– herb steinberg
Nov 19 at 16:50
$frac{a^k-a^{k+1}}{1-a^{n+1}}=a^kfrac{1-a}{1-a^{n+1}}=frac{a^k}{sum_{j=0}^na^j}$, so we agree. (Almost, check lower limit of sum).
– herb steinberg
Nov 19 at 16:50
Since there is a jump at x=0, you need to include k=0 in the sum.
– herb steinberg
Nov 19 at 22:52
Since there is a jump at x=0, you need to include k=0 in the sum.
– herb steinberg
Nov 19 at 22:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
You use a term (probability function) that I am not familiar with. Do you mean probability density function? If so, you are dealing with a discrete distribution with $p_kgt 0$ defined for $x=k, 0le kle n$. $p_k=frac{a^k-a^{k+1}}{1-a^{n+1}}$
answered Nov 18 at 22:19
herb steinberg
2,2732310
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You use a term (probability function) that I am not familiar with. Do you mean probability density function? If so, you are dealing with a discrete distribution with $p_kgt 0$ defined for $x=k, 0le kle n$. $p_k=frac{a^k-a^{k+1}}{1-a^{n+1}}$
answered Nov 18 at 22:19
herb steinberg
2,2732310
add a comment |
up vote
0
down vote
accepted
You use a term (probability function) that I am not familiar with. Do you mean probability density function? If so, you are dealing with a discrete distribution with $p_kgt 0$ defined for $x=k, 0le kle n$. $p_k=frac{a^k-a^{k+1}}{1-a^{n+1}}$
answered Nov 18 at 22:19
herb steinberg
2,2732310
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You use a term (probability function) that I am not familiar with. Do you mean probability density function? If so, you are dealing with a discrete distribution with $p_kgt 0$ defined for $x=k, 0le kle n$. $p_k=frac{a^k-a^{k+1}}{1-a^{n+1}}$
answered Nov 18 at 22:19
herb steinberg
2,2732310
You use a term (probability function) that I am not familiar with. Do you mean probability density function? If so, you are dealing with a discrete distribution with $p_kgt 0$ defined for $x=k, 0le kle n$. $p_k=frac{a^k-a^{k+1}}{1-a^{n+1}}$
answered Nov 18 at 22:19
herb steinberg
2,2732310
answered Nov 18 at 22:19
herb steinberg
2,2732310
answered Nov 18 at 22:19
herb steinberg
2,2732310
answered Nov 18 at 22:19
herb steinberg
2,2732310
2,2732310
add a comment |
add a comment |
$F(x)$ has the properties of a distribution function as long as $0lt alt 1$. Define what you are looking for. (Probability function?)
– herb steinberg
Nov 18 at 20:13
Yes, Sorry I forgot to write that: 0 < a < 1. I need to find the Probability function of the corresponding cdf.
– Francisco
Nov 18 at 21:56
My book calls it Probability function. The result is: p(x)= (a^x)/[sum k=1,..,n of (a^k)] for x=0,..,n
– Francisco
Nov 19 at 6:03
$frac{a^k-a^{k+1}}{1-a^{n+1}}=a^kfrac{1-a}{1-a^{n+1}}=frac{a^k}{sum_{j=0}^na^j}$, so we agree. (Almost, check lower limit of sum).
– herb steinberg
Nov 19 at 16:50
Since there is a jump at x=0, you need to include k=0 in the sum.
– herb steinberg
Nov 19 at 22:52