“Following” $operatorname{exp}(lambda)$ random variables “sum” to $operatorname{Poi}(lambda t)$...
$begingroup$
Lifetime of a bulb is distributed $operatorname{exp}(lambda)$. When one light bulb is burned we replace it immidietely. Let $N_t$ be the number of bulbs we've used by time $t$. Prove that $N_t sim operatorname{Poi}(lambda t)$.
Where is the mistake in my solution?
For all $iin mathbb{N}$, let $X_i simoperatorname{exp}(lambda)$ be the lifetime of bulb #$i$. $(X_i)$ are independent.
$$
\ mathbb{P}(N_t=k)=mathbb{P}(t-1<X_1+...+X_kleq t)
\= dots = (tlambda)^{k-1}cdot e^{-lambda t} (e^{-lambda}-1)
$$
probability probability-distributions random-variables poisson-distribution exponential-distribution
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
$endgroup$
add a comment |
$begingroup$
Lifetime of a bulb is distributed $operatorname{exp}(lambda)$. When one light bulb is burned we replace it immidietely. Let $N_t$ be the number of bulbs we've used by time $t$. Prove that $N_t sim operatorname{Poi}(lambda t)$.
Where is the mistake in my solution?
For all $iin mathbb{N}$, let $X_i simoperatorname{exp}(lambda)$ be the lifetime of bulb #$i$. $(X_i)$ are independent.
$$
\ mathbb{P}(N_t=k)=mathbb{P}(t-1<X_1+...+X_kleq t)
\= dots = (tlambda)^{k-1}cdot e^{-lambda t} (e^{-lambda}-1)
$$
probability probability-distributions random-variables poisson-distribution exponential-distribution
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
$endgroup$
add a comment |
$begingroup$
Lifetime of a bulb is distributed $operatorname{exp}(lambda)$. When one light bulb is burned we replace it immidietely. Let $N_t$ be the number of bulbs we've used by time $t$. Prove that $N_t sim operatorname{Poi}(lambda t)$.
Where is the mistake in my solution?
For all $iin mathbb{N}$, let $X_i simoperatorname{exp}(lambda)$ be the lifetime of bulb #$i$. $(X_i)$ are independent.
$$
\ mathbb{P}(N_t=k)=mathbb{P}(t-1<X_1+...+X_kleq t)
\= dots = (tlambda)^{k-1}cdot e^{-lambda t} (e^{-lambda}-1)
$$
probability probability-distributions random-variables poisson-distribution exponential-distribution
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
$endgroup$
Lifetime of a bulb is distributed $operatorname{exp}(lambda)$. When one light bulb is burned we replace it immidietely. Let $N_t$ be the number of bulbs we've used by time $t$. Prove that $N_t sim operatorname{Poi}(lambda t)$.
Where is the mistake in my solution?
For all $iin mathbb{N}$, let $X_i simoperatorname{exp}(lambda)$ be the lifetime of bulb #$i$. $(X_i)$ are independent.
$$
\ mathbb{P}(N_t=k)=mathbb{P}(t-1<X_1+...+X_kleq t)
\= dots = (tlambda)^{k-1}cdot e^{-lambda t} (e^{-lambda}-1)
$$
probability probability-distributions random-variables poisson-distribution exponential-distribution
probability probability-distributions random-variables poisson-distribution exponential-distribution
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
edited Dec 14 '18 at 16:17
J. Doe
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
asked Dec 13 '18 at 14:50
J. DoeJ. Doe
16111
16111
add a comment |
add a comment |
1 Answer
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$begingroup$
Mistakes:
1) In first line on RHS you use $n$ instead of $k$.
2) ${N_t=k}neq{t-1<X_1+cdots+X_kleq t}$
What we do have is ${N_t= k}={X_1+cdots+X_kleq t<X_1+cdots+X_k+X_{k+1}}$.
It is handsome to make use of: $$P(N_t=k)=P(N_tgeq k)-P(N_tgeq k+1)$$
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
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1 Answer
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1 Answer
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$begingroup$
Mistakes:
1) In first line on RHS you use $n$ instead of $k$.
2) ${N_t=k}neq{t-1<X_1+cdots+X_kleq t}$
What we do have is ${N_t= k}={X_1+cdots+X_kleq t<X_1+cdots+X_k+X_{k+1}}$.
It is handsome to make use of: $$P(N_t=k)=P(N_tgeq k)-P(N_tgeq k+1)$$
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Mistakes:
1) In first line on RHS you use $n$ instead of $k$.
2) ${N_t=k}neq{t-1<X_1+cdots+X_kleq t}$
What we do have is ${N_t= k}={X_1+cdots+X_kleq t<X_1+cdots+X_k+X_{k+1}}$.
It is handsome to make use of: $$P(N_t=k)=P(N_tgeq k)-P(N_tgeq k+1)$$
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Mistakes:
1) In first line on RHS you use $n$ instead of $k$.
2) ${N_t=k}neq{t-1<X_1+cdots+X_kleq t}$
What we do have is ${N_t= k}={X_1+cdots+X_kleq t<X_1+cdots+X_k+X_{k+1}}$.
It is handsome to make use of: $$P(N_t=k)=P(N_tgeq k)-P(N_tgeq k+1)$$
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
$endgroup$
Mistakes:
1) In first line on RHS you use $n$ instead of $k$.
2) ${N_t=k}neq{t-1<X_1+cdots+X_kleq t}$
What we do have is ${N_t= k}={X_1+cdots+X_kleq t<X_1+cdots+X_k+X_{k+1}}$.
It is handsome to make use of: $$P(N_t=k)=P(N_tgeq k)-P(N_tgeq k+1)$$
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
answered Dec 13 '18 at 15:07
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
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