Distribution Function Conditioned On Geometric Random Variable
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Let $X’_is$ be independent random variables with some common distribution function $F$, and suppose they are independent of$N$, a geometric random variable with parameter $p$. If $M = max(X_1,X_2,dots,X_n)$, obtain the distribution function of $M$ while conditioning on $N$.
My approach :
Since $M$ is independent of $N$, I just divided the distribution function of $M$ by $P(X=n)$
So I get,
$$F_{M|N}(m|n) = frac{[F_X(a)]^n}{(1-p)^{n-1}p} $$
I am not sure if this is the right method and answer, can someone clarify and provide the correct method if this is wrong.
probability probability-distributions random-variables conditional-probability
asked Dec 3 '18 at 21:38
user601297user601297
37119
$endgroup$
add a comment |
$begingroup$
Let $X’_is$ be independent random variables with some common distribution function $F$, and suppose they are independent of$N$, a geometric random variable with parameter $p$. If $M = max(X_1,X_2,dots,X_n)$, obtain the distribution function of $M$ while conditioning on $N$.
My approach :
Since $M$ is independent of $N$, I just divided the distribution function of $M$ by $P(X=n)$
So I get,
$$F_{M|N}(m|n) = frac{[F_X(a)]^n}{(1-p)^{n-1}p} $$
I am not sure if this is the right method and answer, can someone clarify and provide the correct method if this is wrong.
probability probability-distributions random-variables conditional-probability
asked Dec 3 '18 at 21:38
user601297user601297
37119
$endgroup$
$begingroup$
Do you mean $M = max(X_1,dots,X_N)$? Because otherwise $M$ is independent of $N$.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:29
$begingroup$
@TkiDeneb yes i mean this
$endgroup$
– user601297
Dec 5 '18 at 0:05
$begingroup$
"Since "$M$ is independent of $N$..." $M$ can't be independent of $N$. The larger the sample size, the larger the observed values of $M$ will be. So $N$ affects the distribution of $M$..
$endgroup$
– JimB
Dec 5 '18 at 3:39
$begingroup$
I suspect you are looking for the marginal cumulative distribution function of $M$ rather than the distribution of $M$ conditioned on $N$. To do that you'll need to sum over the values of $F(x|N=n)*Pr(N=n)$ for $n=1$ to $n=infty$. And there are some assumptions of $F$ being absolutely continuous.
$endgroup$
– JimB
Dec 5 '18 at 5:02
add a comment |
$begingroup$
Let $X’_is$ be independent random variables with some common distribution function $F$, and suppose they are independent of$N$, a geometric random variable with parameter $p$. If $M = max(X_1,X_2,dots,X_n)$, obtain the distribution function of $M$ while conditioning on $N$.
My approach :
Since $M$ is independent of $N$, I just divided the distribution function of $M$ by $P(X=n)$
So I get,
$$F_{M|N}(m|n) = frac{[F_X(a)]^n}{(1-p)^{n-1}p} $$
I am not sure if this is the right method and answer, can someone clarify and provide the correct method if this is wrong.
probability probability-distributions random-variables conditional-probability
asked Dec 3 '18 at 21:38
user601297user601297
37119
$endgroup$
Let $X’_is$ be independent random variables with some common distribution function $F$, and suppose they are independent of$N$, a geometric random variable with parameter $p$. If $M = max(X_1,X_2,dots,X_n)$, obtain the distribution function of $M$ while conditioning on $N$.
My approach :
Since $M$ is independent of $N$, I just divided the distribution function of $M$ by $P(X=n)$
So I get,
$$F_{M|N}(m|n) = frac{[F_X(a)]^n}{(1-p)^{n-1}p} $$
I am not sure if this is the right method and answer, can someone clarify and provide the correct method if this is wrong.
probability probability-distributions random-variables conditional-probability
probability probability-distributions random-variables conditional-probability
asked Dec 3 '18 at 21:38
user601297user601297
37119
asked Dec 3 '18 at 21:38
user601297user601297
37119
edited Dec 5 '18 at 0:04
user601297
asked Dec 3 '18 at 21:38
user601297user601297
37119
asked Dec 3 '18 at 21:38
user601297user601297
37119
asked Dec 3 '18 at 21:38
user601297user601297
37119
37119
$begingroup$
Do you mean $M = max(X_1,dots,X_N)$? Because otherwise $M$ is independent of $N$.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:29
$begingroup$
@TkiDeneb yes i mean this
$endgroup$
– user601297
Dec 5 '18 at 0:05
$begingroup$
"Since "$M$ is independent of $N$..." $M$ can't be independent of $N$. The larger the sample size, the larger the observed values of $M$ will be. So $N$ affects the distribution of $M$..
$endgroup$
– JimB
Dec 5 '18 at 3:39
$begingroup$
I suspect you are looking for the marginal cumulative distribution function of $M$ rather than the distribution of $M$ conditioned on $N$. To do that you'll need to sum over the values of $F(x|N=n)*Pr(N=n)$ for $n=1$ to $n=infty$. And there are some assumptions of $F$ being absolutely continuous.
$endgroup$
– JimB
Dec 5 '18 at 5:02
add a comment |
$begingroup$
Do you mean $M = max(X_1,dots,X_N)$? Because otherwise $M$ is independent of $N$.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:29
$begingroup$
@TkiDeneb yes i mean this
$endgroup$
– user601297
Dec 5 '18 at 0:05
$begingroup$
"Since "$M$ is independent of $N$..." $M$ can't be independent of $N$. The larger the sample size, the larger the observed values of $M$ will be. So $N$ affects the distribution of $M$..
$endgroup$
– JimB
Dec 5 '18 at 3:39
$begingroup$
I suspect you are looking for the marginal cumulative distribution function of $M$ rather than the distribution of $M$ conditioned on $N$. To do that you'll need to sum over the values of $F(x|N=n)*Pr(N=n)$ for $n=1$ to $n=infty$. And there are some assumptions of $F$ being absolutely continuous.
$endgroup$
– JimB
Dec 5 '18 at 5:02
$begingroup$
Do you mean $M = max(X_1,dots,X_N)$? Because otherwise $M$ is independent of $N$.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:29
$begingroup$
Do you mean $M = max(X_1,dots,X_N)$? Because otherwise $M$ is independent of $N$.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:29
$begingroup$
@TkiDeneb yes i mean this
$endgroup$
– user601297
Dec 5 '18 at 0:05
$begingroup$
@TkiDeneb yes i mean this
$endgroup$
– user601297
Dec 5 '18 at 0:05
$begingroup$
"Since "$M$ is independent of $N$..." $M$ can't be independent of $N$. The larger the sample size, the larger the observed values of $M$ will be. So $N$ affects the distribution of $M$..
$endgroup$
– JimB
Dec 5 '18 at 3:39
$begingroup$
"Since "$M$ is independent of $N$..." $M$ can't be independent of $N$. The larger the sample size, the larger the observed values of $M$ will be. So $N$ affects the distribution of $M$..
$endgroup$
– JimB
Dec 5 '18 at 3:39
$begingroup$
I suspect you are looking for the marginal cumulative distribution function of $M$ rather than the distribution of $M$ conditioned on $N$. To do that you'll need to sum over the values of $F(x|N=n)*Pr(N=n)$ for $n=1$ to $n=infty$. And there are some assumptions of $F$ being absolutely continuous.
$endgroup$
– JimB
Dec 5 '18 at 5:02
$begingroup$
I suspect you are looking for the marginal cumulative distribution function of $M$ rather than the distribution of $M$ conditioned on $N$. To do that you'll need to sum over the values of $F(x|N=n)*Pr(N=n)$ for $n=1$ to $n=infty$. And there are some assumptions of $F$ being absolutely continuous.
$endgroup$
– JimB
Dec 5 '18 at 5:02
add a comment |
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$begingroup$
Do you mean $M = max(X_1,dots,X_N)$? Because otherwise $M$ is independent of $N$.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:29
$begingroup$
@TkiDeneb yes i mean this
$endgroup$
– user601297
Dec 5 '18 at 0:05
$begingroup$
"Since "$M$ is independent of $N$..." $M$ can't be independent of $N$. The larger the sample size, the larger the observed values of $M$ will be. So $N$ affects the distribution of $M$..
$endgroup$
– JimB
Dec 5 '18 at 3:39
$begingroup$
I suspect you are looking for the marginal cumulative distribution function of $M$ rather than the distribution of $M$ conditioned on $N$. To do that you'll need to sum over the values of $F(x|N=n)*Pr(N=n)$ for $n=1$ to $n=infty$. And there are some assumptions of $F$ being absolutely continuous.
$endgroup$
– JimB
Dec 5 '18 at 5:02