Question

Let $X$ and $Y$ be random variables (on some unspecified probability space $(\Omega, \mathcal{F}, \mathbb{P})$ ), assume they have a joint density $f_{X, Y}(x, y)$, and assume $\mathbb{E}|Y|<\infty$. In particular, for every Borel subset $C$ of $\mathbb{R}^2$, we have $$ \mathbb{P}\{(X, Y) \in C\}=\int_C f_{X, Y}(x, y) d x d y . $$ In elementary probability, one learns to compute $\mathbb{E}[Y \mid X=x]$, which is a nonrandom function of the dummy variable $x$, by the formula $$ \mathbb{E}[Y \mid X=x]=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y, $$ where $f_{Y \mid X}(y \mid x)$ is the conditional density defined by $$ f_{Y \mid X}(y \mid x)=\frac{f_{X, Y}(x, y)}{f_X(x)} . $$ The denominator in this expression, $f_X(x)=\int_{-\infty}^{\infty} f_{X, Y}(x, \eta) d \eta$, is the marginal density of $X$, and we must assume it is strictly positive for every $x$. We introduce the symbol $g(x)$ for the function $\mathbb{E}[Y \mid X=x]$ defined by (2.6.1); i.e., $$ g(x)=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y=\int_{-\infty}^{\infty} \frac{y f_{X, Y}(x, y)}{f_X(x)} d y . $$ In measure-theoretic probability, conditional expectation is a random variable $\mathbb{E}[Y \mid X]$. This exercise is to show that when there is a joint density for $(X, Y)$, this random variable can be obtained by substituting the random variable $X$ in place of the dummy variable $x$ in the function $g(x)$. In other words, this exercise is to show that $$ \mathbf{E}[Y \mid X]=g(X) . $$ (We introduced the symbol $g(x)$ in order to avoid the mathematically confusing expression $E[Y \mid X=X]$.)

     Let $X$ and $Y$ be random variables (on some unspecified probability space $(\Omega, \mathcal{F}, \mathbb{P})$ ), assume they have a joint density $f_{X, Y}(x, y)$, and assume $\mathbb{E}|Y|<\infty$. In particular, for every Borel subset $C$ of $\mathbb{R}^2$, we have
$$
\mathbb{P}\{(X, Y) \in C\}=\int_C f_{X, Y}(x, y) d x d y .
$$

In elementary probability, one learns to compute $\mathbb{E}[Y \mid X=x]$, which is a nonrandom function of the dummy variable $x$, by the formula
$$
\mathbb{E}[Y \mid X=x]=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y,
$$
where $f_{Y \mid X}(y \mid x)$ is the conditional density defined by
$$
f_{Y \mid X}(y \mid x)=\frac{f_{X, Y}(x, y)}{f_X(x)} .
$$

The denominator in this expression, $f_X(x)=\int_{-\infty}^{\infty} f_{X, Y}(x, \eta) d \eta$, is the marginal density of $X$, and we must assume it is strictly positive for every $x$. We introduce the symbol $g(x)$ for the function $\mathbb{E}[Y \mid X=x]$ defined by (2.6.1); i.e.,
$$
g(x)=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y=\int_{-\infty}^{\infty} \frac{y f_{X, Y}(x, y)}{f_X(x)} d y .
$$

In measure-theoretic probability, conditional expectation is a random variable $\mathbb{E}[Y \mid X]$. This exercise is to show that when there is a joint density for $(X, Y)$, this random variable can be obtained by substituting the random variable $X$ in place of the dummy variable $x$ in the function $g(x)$. In other words, this exercise is to show that
$$
\mathbf{E}[Y \mid X]=g(X) .
$$
(We introduced the symbol $g(x)$ in order to avoid the mathematically confusing expression $E[Y \mid X=X]$.)
Show more…
Stochastic Calculus for Finance II : Continuous-Time Models
Stochastic Calculus for Finance II : Continuous-Time Models
Steven E. Shreve 1st Edition
Chapter 2, Problem 10 ↓

Instant Answer

verified

Step 1

It satisfies the property that for any measurable function $h(X)$, $$ \mathbb{E}[h(X) \mathbb{E}[Y \mid X]] = \mathbb{E}[h(X) Y]. $$ This property is known as the tower property or the law of iterated expectation.  Show more…

Show all steps

lock
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Let $X$ and $Y$ be random variables (on some unspecified probability space $(\Omega, \mathcal{F}, \mathbb{P})$ ), assume they have a joint density $f_{X, Y}(x, y)$, and assume $\mathbb{E}|Y|<\infty$. In particular, for every Borel subset $C$ of $\mathbb{R}^2$, we have $$ \mathbb{P}\{(X, Y) \in C\}=\int_C f_{X, Y}(x, y) d x d y . $$ In elementary probability, one learns to compute $\mathbb{E}[Y \mid X=x]$, which is a nonrandom function of the dummy variable $x$, by the formula $$ \mathbb{E}[Y \mid X=x]=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y, $$ where $f_{Y \mid X}(y \mid x)$ is the conditional density defined by $$ f_{Y \mid X}(y \mid x)=\frac{f_{X, Y}(x, y)}{f_X(x)} . $$ The denominator in this expression, $f_X(x)=\int_{-\infty}^{\infty} f_{X, Y}(x, \eta) d \eta$, is the marginal density of $X$, and we must assume it is strictly positive for every $x$. We introduce the symbol $g(x)$ for the function $\mathbb{E}[Y \mid X=x]$ defined by (2.6.1); i.e., $$ g(x)=\int_{-\infty}^{\infty} y f_{Y \mid X}(y \mid x) d y=\int_{-\infty}^{\infty} \frac{y f_{X, Y}(x, y)}{f_X(x)} d y . $$ In measure-theoretic probability, conditional expectation is a random variable $\mathbb{E}[Y \mid X]$. This exercise is to show that when there is a joint density for $(X, Y)$, this random variable can be obtained by substituting the random variable $X$ in place of the dummy variable $x$ in the function $g(x)$. In other words, this exercise is to show that $$ \mathbf{E}[Y \mid X]=g(X) . $$ (We introduced the symbol $g(x)$ in order to avoid the mathematically confusing expression $E[Y \mid X=X]$.)
Close icon
Play audio
Feedback
Powered by NumerAI
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever