Find the probability function of random variable
Let $Nsimtext{poisson}(lambda)$ and let $h(x)=e^{ux}$ for some $u in mathbb{R}$ and $u$ is not zero , let $M=h circ N$ $$ $$I want to compute the probability function of $M$ ,
my attempt so far:
$$ P_M(m)= P(M=m)=P(e^{un}=m) =P(un=ln(m)=P(N=frac{ln(m)}{u}) $$
I want to say that if $dfrac{ln(m)}{u} in mathbb{N}$ then $$P_M(m) = e^{-lambda} cdot frac{lambda^frac{ln(m)}{u}}{(frac{ln(m)}{u})!} $$
But i am not sure if what i am saying is true, because i know what happens when $P(N=m)$ and not when $P(N=frac{ln(m)}{u})$
probability
edited Dec 9 at 16:44
Nosrati
26.5k62353
asked Dec 9 at 15:55
Moshe Levy
315
add a comment |
Let $Nsimtext{poisson}(lambda)$ and let $h(x)=e^{ux}$ for some $u in mathbb{R}$ and $u$ is not zero , let $M=h circ N$ $$ $$I want to compute the probability function of $M$ ,
my attempt so far:
$$ P_M(m)= P(M=m)=P(e^{un}=m) =P(un=ln(m)=P(N=frac{ln(m)}{u}) $$
I want to say that if $dfrac{ln(m)}{u} in mathbb{N}$ then $$P_M(m) = e^{-lambda} cdot frac{lambda^frac{ln(m)}{u}}{(frac{ln(m)}{u})!} $$
But i am not sure if what i am saying is true, because i know what happens when $P(N=m)$ and not when $P(N=frac{ln(m)}{u})$
probability
edited Dec 9 at 16:44
Nosrati
26.5k62353
asked Dec 9 at 15:55
Moshe Levy
315
What you did is correct. Said a bit differently : if $m=e^{ku}$ for $kin mathbb N$, then $mathbb P{M=m}=e^{-lambda }frac{lambda ^{ln(m)/u}}{(ln(m)/u)!}$ and $0$ otherwise.
– Surb
Dec 9 at 16:12
add a comment |
Let $Nsimtext{poisson}(lambda)$ and let $h(x)=e^{ux}$ for some $u in mathbb{R}$ and $u$ is not zero , let $M=h circ N$ $$ $$I want to compute the probability function of $M$ ,
my attempt so far:
$$ P_M(m)= P(M=m)=P(e^{un}=m) =P(un=ln(m)=P(N=frac{ln(m)}{u}) $$
I want to say that if $dfrac{ln(m)}{u} in mathbb{N}$ then $$P_M(m) = e^{-lambda} cdot frac{lambda^frac{ln(m)}{u}}{(frac{ln(m)}{u})!} $$
But i am not sure if what i am saying is true, because i know what happens when $P(N=m)$ and not when $P(N=frac{ln(m)}{u})$
probability
edited Dec 9 at 16:44
Nosrati
26.5k62353
asked Dec 9 at 15:55
Moshe Levy
315
Let $Nsimtext{poisson}(lambda)$ and let $h(x)=e^{ux}$ for some $u in mathbb{R}$ and $u$ is not zero , let $M=h circ N$ $$ $$I want to compute the probability function of $M$ ,
my attempt so far:
$$ P_M(m)= P(M=m)=P(e^{un}=m) =P(un=ln(m)=P(N=frac{ln(m)}{u}) $$
I want to say that if $dfrac{ln(m)}{u} in mathbb{N}$ then $$P_M(m) = e^{-lambda} cdot frac{lambda^frac{ln(m)}{u}}{(frac{ln(m)}{u})!} $$
But i am not sure if what i am saying is true, because i know what happens when $P(N=m)$ and not when $P(N=frac{ln(m)}{u})$
probability
probability
edited Dec 9 at 16:44
Nosrati
26.5k62353
asked Dec 9 at 15:55
Moshe Levy
315
edited Dec 9 at 16:44
Nosrati
26.5k62353
asked Dec 9 at 15:55
Moshe Levy
315
edited Dec 9 at 16:44
Nosrati
26.5k62353
edited Dec 9 at 16:44
Nosrati
26.5k62353
edited Dec 9 at 16:44
Nosrati
26.5k62353
26.5k62353
asked Dec 9 at 15:55
Moshe Levy
315
asked Dec 9 at 15:55
Moshe Levy
315
asked Dec 9 at 15:55
Moshe Levy
315
315
What you did is correct. Said a bit differently : if $m=e^{ku}$ for $kin mathbb N$, then $mathbb P{M=m}=e^{-lambda }frac{lambda ^{ln(m)/u}}{(ln(m)/u)!}$ and $0$ otherwise.
– Surb
Dec 9 at 16:12
add a comment |
What you did is correct. Said a bit differently : if $m=e^{ku}$ for $kin mathbb N$, then $mathbb P{M=m}=e^{-lambda }frac{lambda ^{ln(m)/u}}{(ln(m)/u)!}$ and $0$ otherwise.
– Surb
Dec 9 at 16:12
What you did is correct. Said a bit differently : if $m=e^{ku}$ for $kin mathbb N$, then $mathbb P{M=m}=e^{-lambda }frac{lambda ^{ln(m)/u}}{(ln(m)/u)!}$ and $0$ otherwise.
– Surb
Dec 9 at 16:12
What you did is correct. Said a bit differently : if $m=e^{ku}$ for $kin mathbb N$, then $mathbb P{M=m}=e^{-lambda }frac{lambda ^{ln(m)/u}}{(ln(m)/u)!}$ and $0$ otherwise.
– Surb
Dec 9 at 16:12
add a comment |
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What you did is correct. Said a bit differently : if $m=e^{ku}$ for $kin mathbb N$, then $mathbb P{M=m}=e^{-lambda }frac{lambda ^{ln(m)/u}}{(ln(m)/u)!}$ and $0$ otherwise.
– Surb
Dec 9 at 16:12