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Finding the distribution of a random variable
Find the distribution of $overlineY$.convergence in law of Cauchy random variables.Truncated random variablesProve symmetry of probabilities given random variables are iid and have continuous cdfProbability Theory - Transformation of independent continuous random variablesCan I assume that random variables with exponential distribution are positive?Series of Bernoulli random variables has geometric distributionConvergence in distribution for two random variablesCompute the value of a cumulative distribution function of a multivariate distribution (with permutations)Find the density of $Y = a/(1 + X^2)$, where $X$ has the Cauchy distribution.
$begingroup$
$textbfProblem$ Let $Y_1,Y_2,cdots$ be independent and identically distributed with
beginalign*
&P[Y_n=0]=alpha\
&P[Y_n>y]=(1-alpha)e^-y
endalign*
Define the random variables $X_n, ngeq 0$ by
beginalign*
&X_0=0\
&X_n+1=alpha X_n+Y_n+1
endalign*
Prove that
beginalign*
&P[X_n=0]=alpha^n\
&P[X_n>x]=(1-alpha^n)e^-x
endalign*
$textbfAttempt$ Firstly, I found $X_n=alpha^n-1Y_1+cdots+alpha Y_n-1+Y_n$.
If $X_n=0$ , then $Y_1=Y_2=cdots =Y_n=0 $ since $Y_i geq 0 $ for all $1leq i leq n$.
Thus, I got $P[X_n=0]=alpha^n$.
However, I stuck how to $P[X_n>x]$ calculate...
Any help is appreciated... Thank you!
$textbfUpdate$ I'll use induction.
If n=1, we can easily check that $P[X_n>x]=(1-alpha^n)e^-x$
beginalign*
P[X_n+1>x] &= P[alpha X_n+Y_n+1>x]\
&=P[Y_n+1=0]P[alpha X_n>x]+int_0^x P[Y_n+1=t] P[alpha X_n>x-t]dt +int_x^infty P[Y_n+1=t]dt \
&(textrmbecause $ alpha X_n +Y_n+1>x $ if $ Y_n+1>x $)\
&=alpha(1-alpha^n)e^-fracxalpha+int_0^x (1-alpha)e^-t (1-alpha^n)e^-fracx-talphadt +int_x^infty (1-alpha)e^-tdt \
&=alpha(1-alpha^n)e^-fracxalpha+alpha(1-alpha^n)(e^-x-e^-fracxalpha)+(1-alpha)e^-x\
&=(1-alpha^n+1) e^-x
endalign*
I'm not sure my proof is right.....
Could you check my error in my proof?
probability probability-theory stochastic-processes
asked yesterday
w.sdkaw.sdka
36919
$endgroup$
|
show 1 more comment
$begingroup$
$textbfProblem$ Let $Y_1,Y_2,cdots$ be independent and identically distributed with
beginalign*
&P[Y_n=0]=alpha\
&P[Y_n>y]=(1-alpha)e^-y
endalign*
Define the random variables $X_n, ngeq 0$ by
beginalign*
&X_0=0\
&X_n+1=alpha X_n+Y_n+1
endalign*
Prove that
beginalign*
&P[X_n=0]=alpha^n\
&P[X_n>x]=(1-alpha^n)e^-x
endalign*
$textbfAttempt$ Firstly, I found $X_n=alpha^n-1Y_1+cdots+alpha Y_n-1+Y_n$.
If $X_n=0$ , then $Y_1=Y_2=cdots =Y_n=0 $ since $Y_i geq 0 $ for all $1leq i leq n$.
Thus, I got $P[X_n=0]=alpha^n$.
However, I stuck how to $P[X_n>x]$ calculate...
Any help is appreciated... Thank you!
$textbfUpdate$ I'll use induction.
If n=1, we can easily check that $P[X_n>x]=(1-alpha^n)e^-x$
beginalign*
P[X_n+1>x] &= P[alpha X_n+Y_n+1>x]\
&=P[Y_n+1=0]P[alpha X_n>x]+int_0^x P[Y_n+1=t] P[alpha X_n>x-t]dt +int_x^infty P[Y_n+1=t]dt \
&(textrmbecause $ alpha X_n +Y_n+1>x $ if $ Y_n+1>x $)\
&=alpha(1-alpha^n)e^-fracxalpha+int_0^x (1-alpha)e^-t (1-alpha^n)e^-fracx-talphadt +int_x^infty (1-alpha)e^-tdt \
&=alpha(1-alpha^n)e^-fracxalpha+alpha(1-alpha^n)(e^-x-e^-fracxalpha)+(1-alpha)e^-x\
&=(1-alpha^n+1) e^-x
endalign*
I'm not sure my proof is right.....
Could you check my error in my proof?
probability probability-theory stochastic-processes
asked yesterday
w.sdkaw.sdka
36919
$endgroup$
1
$begingroup$
I am pretty sure an inductive argument is the one you should look for
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@Stan Tendijck Umm.... Then, we'll show that $P[X_n+1>x ] = (1-alpha^n+1)e^-x$. And, $P[X_n+1>x] = P[alpha X_n + Y_n+1 >x ]$. How to divide $P[alpha X_n +Y_n+1 >x]$??...
$endgroup$
– w.sdka
yesterday
$begingroup$
Do you know how convolution work, i.e., how you determine the distribution of $X+Y$ where $X$ and $Y$ are some random variables?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
I tried to calculate in 'Update'.... Could you see my attempt??
$endgroup$
– w.sdka
yesterday
$begingroup$
Looks right! Maybe worth noticing that you used $P(Y_n+1=t)=P(Y_n+1>t)$ which is true, if you view the left-hand side as the density of $Y_n+1$ at $t$.
$endgroup$
– Stan Tendijck
yesterday
|
show 1 more comment
$begingroup$
$textbfProblem$ Let $Y_1,Y_2,cdots$ be independent and identically distributed with
beginalign*
&P[Y_n=0]=alpha\
&P[Y_n>y]=(1-alpha)e^-y
endalign*
Define the random variables $X_n, ngeq 0$ by
beginalign*
&X_0=0\
&X_n+1=alpha X_n+Y_n+1
endalign*
Prove that
beginalign*
&P[X_n=0]=alpha^n\
&P[X_n>x]=(1-alpha^n)e^-x
endalign*
$textbfAttempt$ Firstly, I found $X_n=alpha^n-1Y_1+cdots+alpha Y_n-1+Y_n$.
If $X_n=0$ , then $Y_1=Y_2=cdots =Y_n=0 $ since $Y_i geq 0 $ for all $1leq i leq n$.
Thus, I got $P[X_n=0]=alpha^n$.
However, I stuck how to $P[X_n>x]$ calculate...
Any help is appreciated... Thank you!
$textbfUpdate$ I'll use induction.
If n=1, we can easily check that $P[X_n>x]=(1-alpha^n)e^-x$
beginalign*
P[X_n+1>x] &= P[alpha X_n+Y_n+1>x]\
&=P[Y_n+1=0]P[alpha X_n>x]+int_0^x P[Y_n+1=t] P[alpha X_n>x-t]dt +int_x^infty P[Y_n+1=t]dt \
&(textrmbecause $ alpha X_n +Y_n+1>x $ if $ Y_n+1>x $)\
&=alpha(1-alpha^n)e^-fracxalpha+int_0^x (1-alpha)e^-t (1-alpha^n)e^-fracx-talphadt +int_x^infty (1-alpha)e^-tdt \
&=alpha(1-alpha^n)e^-fracxalpha+alpha(1-alpha^n)(e^-x-e^-fracxalpha)+(1-alpha)e^-x\
&=(1-alpha^n+1) e^-x
endalign*
I'm not sure my proof is right.....
Could you check my error in my proof?
probability probability-theory stochastic-processes
asked yesterday
w.sdkaw.sdka
36919
$endgroup$
$textbfProblem$ Let $Y_1,Y_2,cdots$ be independent and identically distributed with
beginalign*
&P[Y_n=0]=alpha\
&P[Y_n>y]=(1-alpha)e^-y
endalign*
Define the random variables $X_n, ngeq 0$ by
beginalign*
&X_0=0\
&X_n+1=alpha X_n+Y_n+1
endalign*
Prove that
beginalign*
&P[X_n=0]=alpha^n\
&P[X_n>x]=(1-alpha^n)e^-x
endalign*
$textbfAttempt$ Firstly, I found $X_n=alpha^n-1Y_1+cdots+alpha Y_n-1+Y_n$.
If $X_n=0$ , then $Y_1=Y_2=cdots =Y_n=0 $ since $Y_i geq 0 $ for all $1leq i leq n$.
Thus, I got $P[X_n=0]=alpha^n$.
However, I stuck how to $P[X_n>x]$ calculate...
Any help is appreciated... Thank you!
$textbfUpdate$ I'll use induction.
If n=1, we can easily check that $P[X_n>x]=(1-alpha^n)e^-x$
beginalign*
P[X_n+1>x] &= P[alpha X_n+Y_n+1>x]\
&=P[Y_n+1=0]P[alpha X_n>x]+int_0^x P[Y_n+1=t] P[alpha X_n>x-t]dt +int_x^infty P[Y_n+1=t]dt \
&(textrmbecause $ alpha X_n +Y_n+1>x $ if $ Y_n+1>x $)\
&=alpha(1-alpha^n)e^-fracxalpha+int_0^x (1-alpha)e^-t (1-alpha^n)e^-fracx-talphadt +int_x^infty (1-alpha)e^-tdt \
&=alpha(1-alpha^n)e^-fracxalpha+alpha(1-alpha^n)(e^-x-e^-fracxalpha)+(1-alpha)e^-x\
&=(1-alpha^n+1) e^-x
endalign*
I'm not sure my proof is right.....
Could you check my error in my proof?
probability probability-theory stochastic-processes
probability probability-theory stochastic-processes
asked yesterday
w.sdkaw.sdka
36919
asked yesterday
w.sdkaw.sdka
36919
edited yesterday
w.sdka
asked yesterday
w.sdkaw.sdka
36919
asked yesterday
w.sdkaw.sdka
36919
asked yesterday
w.sdkaw.sdka
36919
36919
1
$begingroup$
I am pretty sure an inductive argument is the one you should look for
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@Stan Tendijck Umm.... Then, we'll show that $P[X_n+1>x ] = (1-alpha^n+1)e^-x$. And, $P[X_n+1>x] = P[alpha X_n + Y_n+1 >x ]$. How to divide $P[alpha X_n +Y_n+1 >x]$??...
$endgroup$
– w.sdka
yesterday
$begingroup$
Do you know how convolution work, i.e., how you determine the distribution of $X+Y$ where $X$ and $Y$ are some random variables?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
I tried to calculate in 'Update'.... Could you see my attempt??
$endgroup$
– w.sdka
yesterday
$begingroup$
Looks right! Maybe worth noticing that you used $P(Y_n+1=t)=P(Y_n+1>t)$ which is true, if you view the left-hand side as the density of $Y_n+1$ at $t$.
$endgroup$
– Stan Tendijck
yesterday
|
show 1 more comment
1
$begingroup$
I am pretty sure an inductive argument is the one you should look for
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@Stan Tendijck Umm.... Then, we'll show that $P[X_n+1>x ] = (1-alpha^n+1)e^-x$. And, $P[X_n+1>x] = P[alpha X_n + Y_n+1 >x ]$. How to divide $P[alpha X_n +Y_n+1 >x]$??...
$endgroup$
– w.sdka
yesterday
$begingroup$
Do you know how convolution work, i.e., how you determine the distribution of $X+Y$ where $X$ and $Y$ are some random variables?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
I tried to calculate in 'Update'.... Could you see my attempt??
$endgroup$
– w.sdka
yesterday
$begingroup$
Looks right! Maybe worth noticing that you used $P(Y_n+1=t)=P(Y_n+1>t)$ which is true, if you view the left-hand side as the density of $Y_n+1$ at $t$.
$endgroup$
– Stan Tendijck
yesterday
1
1
$begingroup$
I am pretty sure an inductive argument is the one you should look for
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
I am pretty sure an inductive argument is the one you should look for
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@Stan Tendijck Umm.... Then, we'll show that $P[X_n+1>x ] = (1-alpha^n+1)e^-x$. And, $P[X_n+1>x] = P[alpha X_n + Y_n+1 >x ]$. How to divide $P[alpha X_n +Y_n+1 >x]$??...
$endgroup$
– w.sdka
yesterday
$begingroup$
@Stan Tendijck Umm.... Then, we'll show that $P[X_n+1>x ] = (1-alpha^n+1)e^-x$. And, $P[X_n+1>x] = P[alpha X_n + Y_n+1 >x ]$. How to divide $P[alpha X_n +Y_n+1 >x]$??...
$endgroup$
– w.sdka
yesterday
$begingroup$
Do you know how convolution work, i.e., how you determine the distribution of $X+Y$ where $X$ and $Y$ are some random variables?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
Do you know how convolution work, i.e., how you determine the distribution of $X+Y$ where $X$ and $Y$ are some random variables?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
I tried to calculate in 'Update'.... Could you see my attempt??
$endgroup$
– w.sdka
yesterday
$begingroup$
I tried to calculate in 'Update'.... Could you see my attempt??
$endgroup$
– w.sdka
yesterday
$begingroup$
Looks right! Maybe worth noticing that you used $P(Y_n+1=t)=P(Y_n+1>t)$ which is true, if you view the left-hand side as the density of $Y_n+1$ at $t$.
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
Looks right! Maybe worth noticing that you used $P(Y_n+1=t)=P(Y_n+1>t)$ which is true, if you view the left-hand side as the density of $Y_n+1$ at $t$.
$endgroup$
– Stan Tendijck
yesterday
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
An alternative strategy notes that $Y_n$ has characteristic function $alpha+frac1-alpha1-it=frac1-alpha it1-it$, so $X_n$ has characteristic function $prod_j=1^nfrac1-alpha^n-j+1it1-ialpha^n-jt=frac1-alpha^nit1-it$ as required.
answered yesterday
J.G.J.G.
29.2k22845
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
An alternative strategy notes that $Y_n$ has characteristic function $alpha+frac1-alpha1-it=frac1-alpha it1-it$, so $X_n$ has characteristic function $prod_j=1^nfrac1-alpha^n-j+1it1-ialpha^n-jt=frac1-alpha^nit1-it$ as required.
answered yesterday
J.G.J.G.
29.2k22845
$endgroup$
add a comment |
$begingroup$
An alternative strategy notes that $Y_n$ has characteristic function $alpha+frac1-alpha1-it=frac1-alpha it1-it$, so $X_n$ has characteristic function $prod_j=1^nfrac1-alpha^n-j+1it1-ialpha^n-jt=frac1-alpha^nit1-it$ as required.
answered yesterday
J.G.J.G.
29.2k22845
$endgroup$
add a comment |
$begingroup$
An alternative strategy notes that $Y_n$ has characteristic function $alpha+frac1-alpha1-it=frac1-alpha it1-it$, so $X_n$ has characteristic function $prod_j=1^nfrac1-alpha^n-j+1it1-ialpha^n-jt=frac1-alpha^nit1-it$ as required.
answered yesterday
J.G.J.G.
29.2k22845
$endgroup$
An alternative strategy notes that $Y_n$ has characteristic function $alpha+frac1-alpha1-it=frac1-alpha it1-it$, so $X_n$ has characteristic function $prod_j=1^nfrac1-alpha^n-j+1it1-ialpha^n-jt=frac1-alpha^nit1-it$ as required.
answered yesterday
J.G.J.G.
29.2k22845
answered yesterday
J.G.J.G.
29.2k22845
answered yesterday
J.G.J.G.
29.2k22845
answered yesterday
J.G.J.G.
29.2k22845
29.2k22845
add a comment |
add a comment |
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1
$begingroup$
I am pretty sure an inductive argument is the one you should look for
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
@Stan Tendijck Umm.... Then, we'll show that $P[X_n+1>x ] = (1-alpha^n+1)e^-x$. And, $P[X_n+1>x] = P[alpha X_n + Y_n+1 >x ]$. How to divide $P[alpha X_n +Y_n+1 >x]$??...
$endgroup$
– w.sdka
yesterday
$begingroup$
Do you know how convolution work, i.e., how you determine the distribution of $X+Y$ where $X$ and $Y$ are some random variables?
$endgroup$
– Stan Tendijck
yesterday
$begingroup$
I tried to calculate in 'Update'.... Could you see my attempt??
$endgroup$
– w.sdka
yesterday
$begingroup$
Looks right! Maybe worth noticing that you used $P(Y_n+1=t)=P(Y_n+1>t)$ which is true, if you view the left-hand side as the density of $Y_n+1$ at $t$.
$endgroup$
– Stan Tendijck
yesterday