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How can I prove that the function $$\sigma_i\left(t\right) = k_i\left[\left(a+b\left(T_i-t\right)\right)e^{-c\left(T_i-t\right)}+d\right]$$ is bounded/unbounded?

$\sigma_i\left(t\right)$ is the chosen volatility parametrization in the Libor rate dynamics

$$dL_i\left(t\right)=\mu_i\left(t\right)L_i\left(t\right)dt+\sigma_i\left(t\right)L_i\left(t\right)dW_i\left(t\right)$$

I have no clue how to start, any help is appreciated. Thanks in advance.

Edit:

The instantaneous volatility can be decomposed in the following parts $$\sigma_i\left(t\right) = g\left(T_i\right)f\left(T_i-t\right)$$ where $g\left(T_i\right)=k_i$ is the component specific to the individual forward Libor rate and $f\left(T_i-t\right)=\left(a+b\left(T_i-t\right)\right)e^{-c\left(T_i-t\right)}+d$ is the component depending on the residual maturity $T_i-t$.

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  • $\begingroup$ Can you specify what $\phi$ is and how it is defined, as I would be tempted to guess that $\phi$ is the cumulative probability function for the Gaussian distribution. $\endgroup$
    – oliversm
    Jun 8, 2016 at 17:38
  • $\begingroup$ @oliversm I edited my question, I hope it is clear now. I have changed $\phi$ into $k$ to avoid confusion. $\endgroup$
    – Tinkerbell
    Jun 9, 2016 at 7:35
  • $\begingroup$ assuming $T_i-t>0$, $c>0$ will be enough $\endgroup$
    – M. Jeunesse
    Jun 9, 2016 at 8:05
  • $\begingroup$ @Tinkerbell, typo: In the libor rate dynamics shouldn't it be $\textrm{d}L_i(t) = \dots$? $\endgroup$
    – oliversm
    Jun 9, 2016 at 9:50

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If I have read the question correctly then I will assume that $a$, $b$, $c$, $d$, $T_i$, and $k_i$ are constants. If this is the case then the only term which we need to show is bounded is $$ \big(a + b(T_i - t)\big)\exp\big(-c(T_i-t)\big). $$ If we assume that we are only considering the temporal domain $0 \leq t \leq T_i$ such that $T_i - t \geq 0 $ then we can ensure that the the product will remain bounded if we have $c>0$ as this will exponentially dampen the product of the two terms.

If though for whatever reason we have the more general case where $ -\infty \leq t \leq \infty$ then it will all depend again on the sign of $c$, but we would only need to consider these two extreme limits.

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