Probability function- Random variable
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
191k27223437
asked Nov 18 at 11:47
Francisco
21
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
191k27223437
asked Nov 18 at 11:47
Francisco
21
$F(x)$ 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
191k27223437
asked Nov 18 at 11:47
Francisco
21
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
191k27223437
asked Nov 18 at 11:47
Francisco
21
edited Nov 18 at 11:58
amWhy
191k27223437
asked Nov 18 at 11:47
Francisco
21
edited Nov 18 at 11:58
amWhy
191k27223437
edited Nov 18 at 11:58
amWhy
191k27223437
edited Nov 18 at 11:58
amWhy
191k27223437
191k27223437
asked Nov 18 at 11:47
Francisco
21
asked Nov 18 at 11:47
Francisco
21
asked Nov 18 at 11:47
Francisco
21
21
$F(x)$ 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
2,2032310
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
2,2032310
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
2,2032310
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
2,2032310
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
2,2032310
answered Nov 18 at 22:19
herb steinberg
2,2032310
answered Nov 18 at 22:19
herb steinberg
2,2032310
answered Nov 18 at 22:19
herb steinberg
2,2032310
2,2032310
add a comment |
add a comment |
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$F(x)$ 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