Expected Bank Account Value Under Normally Distributed Rates

Summary

The document derives the expected value of a bank account when the cumulative short rate is normally distributed. Starting from the bank account’s exponential representation, it applies the moment-generating formula for a normal random variable to show that the account value is lognormally distributed. The expectation is the initial account value multiplied by an exponential term that depends on the mean and variance of the integrated rate; the stated parameters yield a time-dependent exponent of 0.155 times time.

The result illustrates why the expected bank account value is not found by simply exponentiating the mean integrated rate: the variance contributes half its value to the exponent. The calculation assumes the stated normal distribution for the integrated rate and a specified initial account value. It gives a general expression rather than a numerical value at time three, since the initial account balance is not supplied.

Key ideas

• The bank account value equals its initial value multiplied by the exponential of the integrated short rate.
• A normally distributed integrated rate makes the bank account value lognormally distributed.
• The expectation of an exponential normal variable includes a variance adjustment equal to half the variance.
• The resulting expected value depends on the initial balance and the time horizon.

Tags

# Compute value of $\mathbb{E}(B_3)$


# Compute value of $\mathbb{E}(B_3)$












I wonder would anybody tell me how to calculate $\mathbb{E}(B_3)$ Assuming that $\int_0^{t}r_s\,ds\sim N(0.03t,0.25t)$, then is

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I have similar problem solved:

Assuming that $\int_0^t r_s ds \sim N(0.01t, 0.2t)$, then compute the value of $\mathbb{E}(D( 0,1))$.

here is my solution:

$D(t,T):=\frac{B_t}{B_T}=e^{\int_t^T r_s ds}$.

And, as the exponent is a normal distribution, we can apply the following,

## $\mathbb{E} (D(0,1))= e^{μ + \sigma^2 / 2} = e^{-0.01+1(0.2) /2} = e^{0.09}$ .

but what about $\mathbb{E}(B_3)$ from previous problem

## Answer by Kevin (score 1, accepted)

https://quant.stackexchange.com/a/49970

This is rather similar to the solution you mentioned in your question :)

Let $(r_t)$ be the short rate with $\int_0^{t}r_s\mathrm{d}s\sim N(0.03t,0.25t)$ and $B_t$ the value of the bank account. Recall that by definition $\mathrm{d}B_t=r_tB_t\mathrm{d}t$ and thus $B_t=B_0\exp\left(\int_0^t r_s\mathrm{d}s\right)$. Thus, $(B_t)$ is for every time point $t$ log-normally distributed with \begin{align*} \mathbb{E}[B_3] &= B_0\mathbb{E}\left[\exp\left(\int_0^t r_s\mathrm{d}s\right)\right] \\ &= B_0 \mathbb{E}\left[e^{0.03t+\sqrt{0.25t}Z}\right] \\ &= B_0 e^{0.155t}, \end{align*} where $Z\sim N(0,1)$. I used that $\mathbb{E}\left[e^{\mu+\sigma Z}\right]=e^{\mu+\frac{1}{2}\sigma^2}$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.