Finding the joint mgf of two random variables given conditions
$begingroup$
Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$
for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.
I need help solving this problem. So we want to find
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$
I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.
EDIT: We can write
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$
First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$
EDIT 2: Using an integration table, we get
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$
Similarly,
$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$
So, the moment generating function is given by
$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$
But, the exponent of $e$ can be rewritten as
$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$
So, our moment generating function is given by
$$e^{s^2 + t^2}/2 $$
probability probability-theory probability-distributions
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
$endgroup$
add a comment |
$begingroup$
Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$
for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.
I need help solving this problem. So we want to find
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$
I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.
EDIT: We can write
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$
First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$
EDIT 2: Using an integration table, we get
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$
Similarly,
$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$
So, the moment generating function is given by
$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$
But, the exponent of $e$ can be rewritten as
$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$
So, our moment generating function is given by
$$e^{s^2 + t^2}/2 $$
probability probability-theory probability-distributions
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
$endgroup$
$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50
$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– stackofhay42
Dec 16 '18 at 20:18
$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46
$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– stackofhay42
Dec 16 '18 at 20:51
add a comment |
$begingroup$
Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$
for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.
I need help solving this problem. So we want to find
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$
I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.
EDIT: We can write
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$
First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$
EDIT 2: Using an integration table, we get
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$
Similarly,
$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$
So, the moment generating function is given by
$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$
But, the exponent of $e$ can be rewritten as
$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$
So, our moment generating function is given by
$$e^{s^2 + t^2}/2 $$
probability probability-theory probability-distributions
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
$endgroup$
Let $X_{1}$ and $X_{2}$ be two independently and identically
distributed random variables with a common
$$f(x) = frac{1}{sqrt{2pi}} cdot text{exp}(-x^{2})$$
for all $x in (-infty, infty)$.
Let $Y_{1} = aX_{1} + bX_{2}$ and $Y_{2} = cX_{1} + dX_{2}$ where parameters $a, b, c, d$ satisfy $a^2 + b^2 = 1$ and $c^2 + d^2 =
1$ and $ac + bd = 0$. Find the joint mgf of $(Y_{1}, Y_{2})$.
I need help solving this problem. So we want to find
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{s(aX_{1} + bX_{2})} cdot e^{t(cX_{1} + dX_{2})}].$$
I don't think we can separate this into the product of two expectations because that would assume $Y_{1}$ and $Y_{2}$ are independent. Also, we can easily find the joint pdf of $X_{1}$ and $X_{2}$, which is just $f(x)^{2}$. I'm just not sure about how to do this problem, and I can't make use of the conditions.
EDIT: We can write
$$mathbb{E}[e^{sY_{1} + tY_{2}}] = mathbb{E}[e^{(sa + tc)X_{1}}] cdot mathbb{E}[e^{(sb + td)X_{2}}].$$
First we compute $mathbb{E}[e^{(sa + tc)X_{1}}]$:
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{1}{sqrt{2pi}} int_{-infty}^{infty} e^{(sa + tc)x_{1} - x_{1}^{2}} mathop{dx_{1}} $$
EDIT 2: Using an integration table, we get
$$mathbb{E}[e^{(sa + tc)X_{1}}] = frac{e^{(sa + tc)^{2}}}{sqrt{2}}$$
Similarly,
$$mathbb{E}[e^{(sb + td)X_{2}}] = frac{e^{(sb + td)^{2}}}{sqrt{2}}$$
So, the moment generating function is given by
$$frac{e^{(sb + td)^{2}}}{sqrt{2}} frac{e^{(sa + tc)^{2}}}{sqrt{2}} = frac{e^{s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2}}{2}$$
But, the exponent of $e$ can be rewritten as
$$s^2b^2 + 2sbtd + t^2d^2 + s^2a^2 + 2satc + t^2c^2 = s^{2}(a^{2} + b^{2}) + t^2(c^2 + d^2) + 2st(ac + bd) = s^2 + t^2. $$
So, our moment generating function is given by
$$e^{s^2 + t^2}/2 $$
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
edited Dec 16 '18 at 20:18
stackofhay42
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
asked Dec 16 '18 at 19:34
stackofhay42stackofhay42
1836
1836
$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50
$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– stackofhay42
Dec 16 '18 at 20:18
$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46
$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– stackofhay42
Dec 16 '18 at 20:51
add a comment |
$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50
$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– stackofhay42
Dec 16 '18 at 20:18
$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46
$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– stackofhay42
Dec 16 '18 at 20:51
$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50
$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50
$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– stackofhay42
Dec 16 '18 at 20:18
$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– stackofhay42
Dec 16 '18 at 20:18
$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46
$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46
$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– stackofhay42
Dec 16 '18 at 20:51
$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– stackofhay42
Dec 16 '18 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can directly use the indepencence of $X_1$ and $X_2$:
$$
Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
$$
Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
$$
Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
$$
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
$endgroup$
add a comment |
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$begingroup$
We can directly use the indepencence of $X_1$ and $X_2$:
$$
Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
$$
Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
$$
Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
$$
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
$endgroup$
add a comment |
$begingroup$
We can directly use the indepencence of $X_1$ and $X_2$:
$$
Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
$$
Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
$$
Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
$$
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
$endgroup$
add a comment |
$begingroup$
We can directly use the indepencence of $X_1$ and $X_2$:
$$
Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
$$
Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
$$
Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
$$
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
$endgroup$
We can directly use the indepencence of $X_1$ and $X_2$:
$$
Eleft[e^{sY_1+tY_2}right] = Eleft[e^{(sa+tc)X_1}e^{(sb+td)X_2}right] = Eleft[e^{(sa+tc)X_1}right]Eleft[e^{(sb+td)X_2}right].
$$
Now for example $(sa+tc)X_1 sim N(0,(sa+tc)^2)$, so the first factor is the mean of a log-normal distribution with parameters $mu = 0$ and $sigma^2 = (sa+tc)^2$, which is
$$
Eleft[e^{(sa+tc)X_1}right] = e^{frac{(sa+tc)^2}{2}}.
$$
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
answered Dec 16 '18 at 21:10
Tki DenebTki Deneb
32710
32710
add a comment |
add a comment |
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$begingroup$
Why don't you write $e^{sY_1 + tY_2} = e^{(sa+tc)X_1}e^{(sb+td)X_2}$ and use the independence?
$endgroup$
– Tki Deneb
Dec 16 '18 at 19:50
$begingroup$
@TkiDeneb think i figured it out ur way. can u check?
$endgroup$
– stackofhay42
Dec 16 '18 at 20:18
$begingroup$
I think it should be $E[e^{(sa+tc)X_1}] = e^{frac{(sa+tc)^2}{2}}$.
$endgroup$
– Tki Deneb
Dec 16 '18 at 20:46
$begingroup$
okay, thanks. i'll check again. BTW, if you post an answer i'll flag you correct
$endgroup$
– stackofhay42
Dec 16 '18 at 20:51