Interview Quesions III

Today’s question will test some of the statistics and correlation I’ve discussed in the last couple of months. Assume throughout that $x\sim {\mathbb N}(0,1)$ and $y\sim {\mathbb N}(0,1)$ are jointly normally distributed such that ${\mathbb E}[x \cdot y] = \rho$

a) Calculate ${\mathbb E}[\ e^x \ ]$
b) Calculate ${\mathbb E}[\ e^x \ | \ y = b\ ]$

The first expectation is of a lognormal variate, and the second is of a lognormal variate conditional on some earlier value of the variate having been a particular value – these are very typical of the sorts of quantities that a quant deals with every day, so the solution will be quite instructive! Before reading the solution have a go at each one, the following posts may be useful: SDEs pt. 1, SDEs pt. 2, Results for Common Distributions

a) Here, we use the standard result for expectations

$\begin{align*} {\mathbb E}[ \ e^x \ ] &= \int^{\infty}_{-\infty} e^x \cdot f(x) \ dx \nonumber \\ \nonumber \\ &= {1 \over \sqrt{2 \pi}}\int^{\infty}_{-\infty} e^x \cdot e^{-{1 \over 2}x^2} \ dx \nonumber \\ \nonumber \\ &= {1 \over \sqrt{2 \pi}}\int^{\infty}_{-\infty} \exp\Bigl( -{1\over 2}\Bigl[ x^2 - 2x + 1 - 1 \Bigr] \Bigr) \ dx \nonumber \\ \nonumber \\ &= {e^{1\over 2} \over \sqrt{2 \pi}}\int^{\infty}_{-\infty} \exp\Bigl( -{1\over 2} (x-1)^2 \Bigr) \ dx \nonumber \\ \nonumber \\ &= e^{1\over 2} \nonumber \end{align}$

b) This one is a little tougher, so first of all I’ll discuss what it means and some possible plans of attack. We want to calculate the expectation of $e^x$ , given that $y$ takes a value of $b$ . Of course, if $x$ and $y$ were independent, this wouldn’t make any difference and the result would be the same. However, because they are correlated, the realised value of $y$ will have an effect on the distribution of $x$ .

To demonstrate this, I’ve plotted a few scatter-graphs illustrating the effect of specifying $y$ on $x$ , with $x$ and $y$ uncorrelated and then becoming increasing more correlated.

When x and y are uncorrelated, the realised value of y doesn't affect the distribution for x, which is still normally distributed around zero — When x and y are uncorrelated, the realised value of y doesn’t affect the distribution for x, which is still normally distributed around zero

When x and y are correlated, the realised value of y has an effect on the distribution of x, which is no longer centered on zero and has a smaller variance

When the correlation of x and y becomes high, the value of x is almost completely determined by y. Now, if y is specified then x is tightly centered around a value far from zero

The simplest way of attempting this calculation is to use the result for jointly normal variates given in an earlier post, which says that if $x$ and $y$ have correlation $\rho$ , we can express $x$ in terms of $y$ and a new variate $z \sim {\mathbb N}(0,1)$ which is uncorrelated with $y$

$x = \rho\cdot y + \sqrt{1 - \rho^2}\cdot z$

$(\ e^x\ |\ y = b\ ) = e^{\rho y + \sqrt{1-\rho^2}z} = e^{\rho b}\cdot e^{\sqrt{1-\rho^2}z}$

Since the value of $y$ is already determined (ie. $y$ = $b$ ), I’ve separated this term out and the only thing I have to calculate is the expectation of the second term in $z$ . Since $y$ and $z$ are independent, we can calculate the expectation of the $z$ , which is the same process as before but featuring slightly more complicated pre-factors

$\begin{align*} {\mathbb E}[ \ e^x \ |\ y = b\ ] &= e^{\rho b} \int^{\infty}_{-\infty} e^{\sqrt{1-\rho^2}z} \cdot f(z) \ dz \nonumber \\ \nonumber \\ &= {e^{\rho b} \over \sqrt{2 \pi}}\int^{\infty}_{-\infty} e^{\sqrt{1-\rho^2}z} \cdot e^{-{1 \over 2}z^2} \ dz \nonumber \\ \nonumber \\ &= {e^{\rho b} \over \sqrt{2 \pi}}\int^{\infty}_{-\infty} \exp\Bigl( -{1\over 2}\Bigl[ z^2 - 2\sqrt{1-\rho^2}z \nonumber \\ & \quad \quad \quad \quad \quad \quad \quad \quad + (1-\rho^2) - (1-\rho^2) \Bigr] \Bigr) \ dz \nonumber \\ \nonumber \\ &= {e^{\rho b} \over \sqrt{2 \pi}} e^{{1 \over 2} (1-\rho^2)} \int^{\infty}_{-\infty} \exp\Bigl( -{1\over 2} \bigl(z-\sqrt{1-\rho^2}\bigr)^2 \Bigr) \ dz \nonumber \\ \nonumber \\ &= e^{{1\over 2}(1-\rho^2)+\rho b} \nonumber \end{align}$

We can check the limiting values of this – if $\rho = 0$ then $x$ and $y$ are independent [this is not a general result by the way – see wikipedia for example – but it IS true for jointly normally distributed variables], in this case ${\mathbb E}[ \ e^x \ |\ y = b\ ] = e^{0.5}$ just as above. If $\rho = \pm 1$ , ${\mathbb E} [\ e^x \ |\ y=b\ ] = e^{\pm b}$ , which also makes sense since in this case $x = \pm y = \pm b$ , so $y$ fully determines the expectation of $x$ .

The more general way to solve this is to use the full 2D joint normal distribution as given in the previous post mentioned before,

$f(x,y) = {1 \over {2\pi \sqrt{1-\rho^2}}} \exp{\Bigl(-{1 \over 2(1-\rho^2)}(x^2 + y^2 - 2\rho xy)\Bigr)}$

This is the joint probability function of $x$ and $y$ , but it’s not quite what we need – the expectation we are trying to calculate is

${\mathbb E}[ \ e^x \ |\ y \ ] = \int^{\infty}_{-\infty} e^x \cdot f(\ x \ |\ y \ ) \ dx$

So we need to calculate the conditional expectation of $x$ given $y$ , for which we need Bayes’ theorem

$f(x,y) = f( \ x \ | \ y \ ) \cdot f(y)$

Putting this together, we have

${\mathbb E}[ \ e^x \ |\ y = b\ ] = \int^{\infty}_{-\infty} e^x \cdot {f(x,y) \over f(y)}\ dx$

$={1 \over {2\pi \sqrt{1-\rho^2}}} \int^{\infty}_{-\infty} { e^x \over e^{-{1\over 2}y^2}} \exp{\Bigl(-{1 \over 2(1-\rho^2)}(x^2 + y^2 - 2\rho xy)\Bigr)} dx$

$={e^{{1\over 2}b^2} \over {2\pi \sqrt{1-\rho^2}}} \int^{\infty}_{-\infty} e^x \exp{\Bigl(-{1 \over 2(1-\rho^2)}(x^2 + b^2 - 2\rho xb)\Bigr)} dx$

This integral is left as an exercise to the reader, but it is very similar to those given above and should give the same answer as the previous expression for ${\mathbb E}[\ e^x \ | \ y = b\ ]$ after some simplification!

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