Product of a Discrete Variable and a Continuous Variable
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How would I go about obtaining the probability density function of a random variable that results in a product of a discrete variable and a continuous variable? I know that if $X$ and $Y$ are both continuous, then the probability density function of $Z=XY$ is given by: $$f_Z(z)=int_{-infty}^{infty}f_X(x)f_Yleft(frac{z}{x}right)frac{1}{|x|}dx$$ but what is the method of obtaining $f_Z(z)$ if $X$ is a discrete variable and $Y$ is continuous?
More specific to my case, I have $X$ following a Rademacher distribution, that is $$begin{align*}
P(X=x) = begin{cases}
frac{1}{2} & x = -1\
frac{1}{2} & x = 1\
0 & otherwise
end{cases}
end{align*}$$
and $Y$ following a Rayleigh distribution. What would be the product of these two random variables?
probability probability-theory probability-distributions random-variables dirac-delta
asked Dec 25 '18 at 13:06
rearea
12
$endgroup$
add a comment |
$begingroup$
How would I go about obtaining the probability density function of a random variable that results in a product of a discrete variable and a continuous variable? I know that if $X$ and $Y$ are both continuous, then the probability density function of $Z=XY$ is given by: $$f_Z(z)=int_{-infty}^{infty}f_X(x)f_Yleft(frac{z}{x}right)frac{1}{|x|}dx$$ but what is the method of obtaining $f_Z(z)$ if $X$ is a discrete variable and $Y$ is continuous?
More specific to my case, I have $X$ following a Rademacher distribution, that is $$begin{align*}
P(X=x) = begin{cases}
frac{1}{2} & x = -1\
frac{1}{2} & x = 1\
0 & otherwise
end{cases}
end{align*}$$
and $Y$ following a Rayleigh distribution. What would be the product of these two random variables?
probability probability-theory probability-distributions random-variables dirac-delta
asked Dec 25 '18 at 13:06
rearea
12
$endgroup$
$begingroup$
Use $f_X(x)=sum_k P(X=k)delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function).
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– J.G.
Dec 25 '18 at 13:13
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@J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous?
$endgroup$
– rea
Dec 25 '18 at 13:35
$begingroup$
Yes, it applies.
$endgroup$
– J.G.
Dec 25 '18 at 14:26
add a comment |
$begingroup$
How would I go about obtaining the probability density function of a random variable that results in a product of a discrete variable and a continuous variable? I know that if $X$ and $Y$ are both continuous, then the probability density function of $Z=XY$ is given by: $$f_Z(z)=int_{-infty}^{infty}f_X(x)f_Yleft(frac{z}{x}right)frac{1}{|x|}dx$$ but what is the method of obtaining $f_Z(z)$ if $X$ is a discrete variable and $Y$ is continuous?
More specific to my case, I have $X$ following a Rademacher distribution, that is $$begin{align*}
P(X=x) = begin{cases}
frac{1}{2} & x = -1\
frac{1}{2} & x = 1\
0 & otherwise
end{cases}
end{align*}$$
and $Y$ following a Rayleigh distribution. What would be the product of these two random variables?
probability probability-theory probability-distributions random-variables dirac-delta
asked Dec 25 '18 at 13:06
rearea
12
$endgroup$
How would I go about obtaining the probability density function of a random variable that results in a product of a discrete variable and a continuous variable? I know that if $X$ and $Y$ are both continuous, then the probability density function of $Z=XY$ is given by: $$f_Z(z)=int_{-infty}^{infty}f_X(x)f_Yleft(frac{z}{x}right)frac{1}{|x|}dx$$ but what is the method of obtaining $f_Z(z)$ if $X$ is a discrete variable and $Y$ is continuous?
More specific to my case, I have $X$ following a Rademacher distribution, that is $$begin{align*}
P(X=x) = begin{cases}
frac{1}{2} & x = -1\
frac{1}{2} & x = 1\
0 & otherwise
end{cases}
end{align*}$$
and $Y$ following a Rayleigh distribution. What would be the product of these two random variables?
probability probability-theory probability-distributions random-variables dirac-delta
probability probability-theory probability-distributions random-variables dirac-delta
asked Dec 25 '18 at 13:06
rearea
12
asked Dec 25 '18 at 13:06
rearea
12
edited Dec 25 '18 at 13:58
rea
asked Dec 25 '18 at 13:06
rearea
12
asked Dec 25 '18 at 13:06
rearea
12
asked Dec 25 '18 at 13:06
rearea
12
12
$begingroup$
Use $f_X(x)=sum_k P(X=k)delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function).
$endgroup$
– J.G.
Dec 25 '18 at 13:13
$begingroup$
@J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous?
$endgroup$
– rea
Dec 25 '18 at 13:35
$begingroup$
Yes, it applies.
$endgroup$
– J.G.
Dec 25 '18 at 14:26
add a comment |
$begingroup$
Use $f_X(x)=sum_k P(X=k)delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function).
$endgroup$
– J.G.
Dec 25 '18 at 13:13
$begingroup$
@J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous?
$endgroup$
– rea
Dec 25 '18 at 13:35
$begingroup$
Yes, it applies.
$endgroup$
– J.G.
Dec 25 '18 at 14:26
$begingroup$
Use $f_X(x)=sum_k P(X=k)delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function).
$endgroup$
– J.G.
Dec 25 '18 at 13:13
$begingroup$
Use $f_X(x)=sum_k P(X=k)delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function).
$endgroup$
– J.G.
Dec 25 '18 at 13:13
$begingroup$
@J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous?
$endgroup$
– rea
Dec 25 '18 at 13:35
$begingroup$
@J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous?
$endgroup$
– rea
Dec 25 '18 at 13:35
$begingroup$
Yes, it applies.
$endgroup$
– J.G.
Dec 25 '18 at 14:26
$begingroup$
Yes, it applies.
$endgroup$
– J.G.
Dec 25 '18 at 14:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can try from the definition:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= sum_{k=1}^infty mathbb{P}[XY le z, X = k] \
&= sum_{k=1}^infty mathbb{P}[Y le z/k] mathbb{P}[X = k] \
&= sum_{k=1}^infty F_Y(z/k) f_X(k).
end{split}
$$
Note I did not handle the case where $k le 0$...
One can alternatively condtion on $Y$ similarly and end up with an integral instead of a sum:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= int_mathbb{R} lim_{epsilon to 0}
mathbb{P}[XY le z, |Y-y| < epsilon] dy \
&= int_mathbb{R} F_X(z/y) f_Y(y) dy.
end{split}
$$
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
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add a comment |
$begingroup$
For the special case of Rademacher $X$, $$P(Zle z)=P(X=1)P(Yle z)+P(X=-1)P(Yge -z)=frac{F_Y(z)+1-F_Y(-z)}{2}.$$with $F_Y$ the cdf of $Y$.
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
We can try from the definition:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= sum_{k=1}^infty mathbb{P}[XY le z, X = k] \
&= sum_{k=1}^infty mathbb{P}[Y le z/k] mathbb{P}[X = k] \
&= sum_{k=1}^infty F_Y(z/k) f_X(k).
end{split}
$$
Note I did not handle the case where $k le 0$...
One can alternatively condtion on $Y$ similarly and end up with an integral instead of a sum:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= int_mathbb{R} lim_{epsilon to 0}
mathbb{P}[XY le z, |Y-y| < epsilon] dy \
&= int_mathbb{R} F_X(z/y) f_Y(y) dy.
end{split}
$$
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
$endgroup$
add a comment |
$begingroup$
We can try from the definition:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= sum_{k=1}^infty mathbb{P}[XY le z, X = k] \
&= sum_{k=1}^infty mathbb{P}[Y le z/k] mathbb{P}[X = k] \
&= sum_{k=1}^infty F_Y(z/k) f_X(k).
end{split}
$$
Note I did not handle the case where $k le 0$...
One can alternatively condtion on $Y$ similarly and end up with an integral instead of a sum:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= int_mathbb{R} lim_{epsilon to 0}
mathbb{P}[XY le z, |Y-y| < epsilon] dy \
&= int_mathbb{R} F_X(z/y) f_Y(y) dy.
end{split}
$$
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
$endgroup$
add a comment |
$begingroup$
We can try from the definition:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= sum_{k=1}^infty mathbb{P}[XY le z, X = k] \
&= sum_{k=1}^infty mathbb{P}[Y le z/k] mathbb{P}[X = k] \
&= sum_{k=1}^infty F_Y(z/k) f_X(k).
end{split}
$$
Note I did not handle the case where $k le 0$...
One can alternatively condtion on $Y$ similarly and end up with an integral instead of a sum:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= int_mathbb{R} lim_{epsilon to 0}
mathbb{P}[XY le z, |Y-y| < epsilon] dy \
&= int_mathbb{R} F_X(z/y) f_Y(y) dy.
end{split}
$$
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
$endgroup$
We can try from the definition:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= sum_{k=1}^infty mathbb{P}[XY le z, X = k] \
&= sum_{k=1}^infty mathbb{P}[Y le z/k] mathbb{P}[X = k] \
&= sum_{k=1}^infty F_Y(z/k) f_X(k).
end{split}
$$
Note I did not handle the case where $k le 0$...
One can alternatively condtion on $Y$ similarly and end up with an integral instead of a sum:
$$
begin{split}
F_Z(z)
&= mathbb{P}[Z le z] \
&= mathbb{P}[XY le z] \
&= int_mathbb{R} lim_{epsilon to 0}
mathbb{P}[XY le z, |Y-y| < epsilon] dy \
&= int_mathbb{R} F_X(z/y) f_Y(y) dy.
end{split}
$$
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
answered Dec 25 '18 at 13:42
gt6989bgt6989b
34.7k22456
34.7k22456
add a comment |
add a comment |
$begingroup$
For the special case of Rademacher $X$, $$P(Zle z)=P(X=1)P(Yle z)+P(X=-1)P(Yge -z)=frac{F_Y(z)+1-F_Y(-z)}{2}.$$with $F_Y$ the cdf of $Y$.
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
$endgroup$
add a comment |
$begingroup$
For the special case of Rademacher $X$, $$P(Zle z)=P(X=1)P(Yle z)+P(X=-1)P(Yge -z)=frac{F_Y(z)+1-F_Y(-z)}{2}.$$with $F_Y$ the cdf of $Y$.
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
$endgroup$
add a comment |
$begingroup$
For the special case of Rademacher $X$, $$P(Zle z)=P(X=1)P(Yle z)+P(X=-1)P(Yge -z)=frac{F_Y(z)+1-F_Y(-z)}{2}.$$with $F_Y$ the cdf of $Y$.
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
$endgroup$
For the special case of Rademacher $X$, $$P(Zle z)=P(X=1)P(Yle z)+P(X=-1)P(Yge -z)=frac{F_Y(z)+1-F_Y(-z)}{2}.$$with $F_Y$ the cdf of $Y$.
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
answered Dec 25 '18 at 14:25
J.G.J.G.
29.4k22846
29.4k22846
add a comment |
add a comment |
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$begingroup$
Use $f_X(x)=sum_k P(X=k)delta (x-k)$ (see en.wikipedia.org/wiki/Dirac_delta_function).
$endgroup$
– J.G.
Dec 25 '18 at 13:13
$begingroup$
@J.G. but does that $f_Z(z)$ formula still apply to this case? Is $f_X(x)$ considered continuous?
$endgroup$
– rea
Dec 25 '18 at 13:35
$begingroup$
Yes, it applies.
$endgroup$
– J.G.
Dec 25 '18 at 14:26